Question 1

A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Accepted Answer

0.5 s. A disturbance on a string travels as a transverse wave, and its speed depends only on the string's tension and how much mass is packed into each metre of it — not on how hard the string was jerked. Step 1: Find the linear mass density. Linear mass density μ is the mass per unit length of the string: μ = m/L where m = 2.50 kg is the mass of the string and L = 20.0 m is its length. μ = (2.50 kg )/(20.0 m ) = 0.125 kg/m Step 2: Find the wave speed. The speed of a transverse wave on a stretched string is v = √ T/(μ) where T = 200 N is the tension in the string. This formula comes from applying Newton's second law to a small element of the string; T supplies the restoring force and μ supplies the inertia — a taut, light string carries the jerk fast, a slack, heavy one carries it slowly. v = √ (200 N )/(0.125 kg/m ) = √ 1600 m^2/ s^2 = 40.0 m/s A step people trip on here: this v is the speed at which the disturbance (the wave pattern) moves along the string, not the speed of any particle of the string itself — the string material only moves up and down. Step 3: Find the time to cross the string. The disturbance travels the full length L of the string at this speed, so the time taken is t = L/v = (20.0 m )/(40.0 m/s ) = 0.500 s All three given quantities (2.50 kg, 200 N, 20.0 m) carry three significant figures, so the answer is kept to three significant figures: 0.500 s. Answer: The disturbance takes 0.500 s to reach the other end of the string.

Question 2

A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s^-1? (g = 9.8 m s^-2 )

Accepted Answer

8.7 s. The stone falls under gravity, and the sound climbs back up at a separate, constant speed — these are two different motions that happen one after the other, and you must add their times, not treat it as one calculation. Step 1 — Time for the stone to fall 300 m. The stone starts from rest and falls freely, so use the second equation of motion for uniformly accelerated motion: h = u t 1 + 1/2g t 1^2 Here u = 0 (dropped, not thrown), h = 300 m is the height of the tower, g = 9.8 m s^-2 is the acceleration due to gravity, and t 1 is the time to reach the water. Since u=0: h = 1/2g t 1^2 ⇒ t 1 = √ 2h/g Substituting values: t 1 = √ (2 × 300 m )/(9.8 m s^-2) = √ 600/9.8 s^2 = √ 61.22 s^2 t 1 = 7.825 s Step 2 — Time for the sound of the splash to travel back up. Once the stone hits the water, the splash produces a sound that travels back up the same 300 m at a constant speed (sound does not accelerate the way a falling body does). Using speed = ( distance )/( time ) ⇒ t 2 = h v sound where v sound = 340 m s^-1. t 2 = (300 m )/(340 m s^-1) = 0.8824 s Step 3 — Add the two times. The splash is heard only after both events are complete: the fall, then the sound's return trip. These are sequential, not simultaneous, so T = t 1 + t 2 = 7.825 s + 0.8824 s = 8.707 s Rounding for significant figures: the given data ( g = 9.8, v = 340 ) carry two significant figures, so the final answer should also be quoted to two significant figures. Rounding 8.707 s to two significant figures gives 8.7 s. Do not round t 1 or t 2 separately before adding — that is a common mistake that can shift the last digit. Answer: The splash is heard at the top 8.7 s after the stone is dropped.

Question 3

A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20^° C = 343 m s^-1.

Accepted Answer

2.06 × 10^4 N. A wave on a stretched wire is governed by its tension and its mass per unit length — set the transverse-wave speed equal to the given speed of sound and solve for the tension. The speed of a transverse wave on a stretched string (or wire) is v = √ T/(μ) where T is the tension in the wire and μ is its linear mass density (mass per unit length). Step 1: Find the linear mass density. μ = m/L = (2.10 kg )/(12.0 m ) = 0.175 kg m^-1 Step 2: Set the wave speed equal to the speed of sound. The wire must carry a transverse wave whose speed matches the speed of sound in dry air at 20^° C, v = 343 m s^-1. This is the speed of the wave on the wire, not the speed of sound traveling through the wire material itself — a distinction worth being careful about, since it is easy to mix up "speed of sound in air" with "speed of sound in the wire." Squaring the wave-speed relation and solving for tension, v^2 = T/(μ) ⇒ T = μ v^2 Step 3: Substitute the numbers. T = (0.175 kg m^-1) × (343 m s^-1)^2 T = 0.175 × 117649 kg m s^-2 T = 20588.575 N Step 4: Round to the correct number of significant figures. Every measured quantity here — 2.10 kg, 12.0 m, and 343 m s^-1 — is given to three significant figures, so the answer must also be quoted to three significant figures, not carried out to five or six digits just because the calculator shows them. T ≈ 2.06 × 10^4 N This is a large tension (about 2100 kgf, or the weight of roughly two tonnes) — a reasonable size for a taut steel wire, since the wave speed required (343 m/s) is very high for a wire this light. Answer: T ≈ 2.06 × 10^4 N (about 20600 N).

Question 4

Use the formula v P = γ ρ to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.

Accepted Answer

Assume ideal gas law: P = ρRT M, where ρ is the density, M is the molecular mass, and T is the temperature of the gas. This gives v = RT M γ. This shows that v is: (a) Independent of pressure. (b) Increases as T. (c) The molecular mass of water (18) is less than that of N 2 (28) and O 2 (32). Therefore as humidity increases, the effective molecular mass of air decreases and hence v increases.. The formula v = √ (γ P)/(ρ) hides temperature and composition inside P and ρ — pull them out using the ideal gas law, and all three parts fall out together. Here v is the speed of sound, γ = C p/C v is the ratio of specific heats of the gas, P is the pressure, and ρ is the density of the gas. For n moles of an ideal gas occupying volume V, the gas law gives PV = nRT where R is the universal gas constant and T is the absolute temperature. If M is the molar mass of the gas and m = nM is the mass of gas in that volume, then the density is ρ = m/V = nM/V, so n/V = ρ/M. Substituting into the gas law, P = n/VRT = (ρ RT)/M ⇒ P/(ρ) = RT/M. Putting this into the speed formula, v = √ (γ P)/(ρ) = √ (γ R T)/M. This rewritten form is the one to use for all three parts, since it separates out T and M explicitly. (a) Independent of pressure. At a fixed temperature, the ratio P/ρ equals RT/M, which does not contain P at all — it depends only on T and on which gas it is. Physically: if you compress air at constant temperature, Boyle's law says P ∝ ρ (density rises in the same proportion as pressure, since the same mass is squeezed into less volume), so the ratio P/ρ stays fixed even though P itself has changed. Since v = √ γ P/ρ depends only on this ratio (and on γ, which is also unchanged for a given gas at a given temperature), v does not change when P alone is changed at constant T. (b) Increases with temperature. From v = √ γ R T/M, for a fixed gas (fixed γ, R, M ), v ∝ √ T. Raising the temperature directly raises v — this is a genuine physical dependence, not a coincidence of the units, because faster-moving, more energetic molecules transmit a pressure disturbance faster. A common slip is to think humidity and temperature act the same way; they don't — temperature raises v by raising T itself in the formula, while humidity (part c) raises v by lowering M. (c) Increases with humidity. Moist air is a mixture of dry air (mean molar mass about 28.9 g/mol, mostly N 2 and O 2 ) with water vapour (molar mass 18 g/mol ) mixed in. By Avogadro's law, equal volumes of gas at the same temperature and pressure contain equal numbers of moles, regardless of what the gas is. So when water vapour molecules replace some of the heavier N 2 / O 2 molecules mole-for-mole, the average molar mass M of the mixture drops below that of dry air, even though T and P are unchanged. Since v = √ (γ R T)/M, and M sits in the denominator, a smaller M gives a larger v. So humid air, being effectively "lighter" per mole than dry air at the same temperature and pressure, carries sound faster than dry air. Answer: (a) At constant temperature P/ρ = RT/M is independent of P (Boyle's law makes ρ rise in step with P ), so v does not depend on pressure. (b) v = √ γ RT/M ∝ √ T, so v increases as temperature rises. (c) Water vapour lowers the mean molar mass M of humid air compared to dry air, and since v ∝ 1/√ M, the speed of sound increases with humidity.

Question 5

You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x - v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave: (a) (x - vt )2 (b) log [(x + vt)/ x 0 ] (c) 1/(x + vt)

Accepted Answer

The converse is not true. An obvious requirement for an acceptable function for a travelling wave is that it should be finite everywhere and at all times. Only function (c) satisfies this condition, the remaining functions cannot possibly represent a travelling wave.. Being a function of x ± vt is necessary for a travelling wave, but it is not sufficient — the function must also stay finite everywhere, at every instant. Start with why the combination x ± vt matters at all. A disturbance that keeps its shape while sliding along the x -axis with speed v must depend on position and time only through x - vt (shape moving in the +x direction) or x + vt (moving in the -x direction). Here y is the displacement of the medium, x the position of a point of the medium, t the time, and v the wave speed. Now the converse. Put u = x ± vt and let y = f(u) be any twice-differentiable function. By the chain rule, (∂ y)/(∂ x) = f'(u), (∂^2 y)/(∂ x^2) = f''(u), (∂ y)/(∂ t) = ± v f'(u), (∂^2 y)/(∂ t^2) = v^2 f''(u), so every such f obeys the wave equation (∂^2 y)/(∂ x^2) = 1/v^2 (∂^2 y)/(∂ t^2). That is exactly why the converse fails as a physical statement: the form f(x± vt) is cheap — all three functions below have it, and all three satisfy the wave equation wherever they are differentiable. What the mathematics does not enforce, and physics does, is this: y is a real displacement of real matter, so it must be - finite (bounded) for every x and every t — no particle of the medium can be pushed aside by an infinite amount; and - defined and single-valued at every x and t. A pulse should also die away, y → 0 as x → ∞, so that the disturbance is localised and carries finite energy. An aside before testing them: strictly, none of the three has the units of a length — (x-vt)^2 is in m^2, 1/(x+vt) in m^-1, and a logarithm needs a dimensionless argument, which is what x 0 is there to supply. Multiplying each by a suitable constant fixes the units and changes nothing about whether the function stays finite, so that is not the issue being tested here. (a) y = (x - vt)^2. It is a function of x - vt, so it passes the form test. But it is a parabola whose minimum sits at x = vt and which grows without limit on either side: at t = 0, y = x^2, so at x = 10 m, y = 100; at x = 100 m, y = 10^4; and y → ∞ as x - vt → ∞. The displacement is unbounded, and the medium is nowhere left undisturbed except at the single point x = vt — the opposite of a pulse. Not an acceptable travelling wave. (b) y = log [(x + vt)/x 0]. Again a function of x + vt, so the form test passes, and it fails everything else: - for x + vt < 0 the argument is negative and the logarithm is not defined at all, so y does not exist over half the axis; - at x + vt = 0, y → -∞; - as x + vt → ∞, y → +∞. Not an acceptable travelling wave. (c) y = 1/(x + vt). This is a function of x + vt, and it is the only one of the three that behaves well far away: y → 0 as x + vt → ∞, so at large distances the medium is undisturbed, which is what a pulse should look like. But look at the point where the denominator vanishes, x + vt = 0, i.e. x = -vt. That is a genuine point of the medium at every instant — for v = 2 m s^-1 and t = 3 s it is the point x = -6 m — and there y → ±∞. The singularity is not at some unreachable place; it travels along the string in the -x direction with speed v, demanding an infinite displacement at whichever point it is passing. An infinite displacement is not physical. Not an acceptable travelling wave either. On the printed key. NCERT states the right criterion — a travelling-wave function must be finite everywhere and at all times — and then concludes that "only function (c) satisfies this condition". That conclusion does not follow from the criterion, and it is wrong: 1/(x+vt) is not finite everywhere, because it diverges on the line x = -vt. Substituting any t shows it. What is true, and is presumably why (c) got singled out, is that (c) is the only one of the three that decays to zero at large distance; but decaying at infinity does not repair a singularity in the middle of the medium. Judged by the very test the book names, all three functions fail. If your teacher or exam follows the printed key, quote the finiteness rule and point at x = -vt — the reasoning, not the key, is what the question is asking for. Answer: No, the converse is not true — depending on x and t only through x ± vt is necessary but not sufficient; the function must in addition remain finite (and defined) for all x and t. None of the three qualifies: (a) (x-vt)^2 grows without bound as x-vt → ∞; (b) log [(x+vt)/x 0] is undefined for x+vt<0 and diverges at x+vt = 0 and as x+vt → ∞; (c) 1/(x+vt), although it decays at large distance, becomes infinite at x = -vt. NCERT's printed answer accepting (c) contradicts the finiteness condition it itself states.

Question 6

A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s^-1 and in water 1486 m s^-1.

Accepted Answer

(a) 3.4 × 10^-4 m (b) 1.49 ×× 10^-3 m. The frequency never changes at a boundary — only the speed does, so the wavelength on each side is different. A bat's ultrasonic call hits the air–water surface and splits into two waves: part of it bounces back into the air (the reflected wave), part of it crosses into the water (the transmitted wave). Both waves are still driven by the same source, so both carry the same frequency as the original 1000 kHz call. What changes from one medium to the other is the speed of sound, and since v = fλ ⇒ λ = v/f a change in speed at fixed frequency forces a change in wavelength. This is the point people get wrong here — it's easy to assume wavelength is "conserved" across a boundary, but it is frequency that stays fixed, not wavelength. Setting up the numbers Convert the frequency to hertz first: f = 1000 kHz = 1000 × 10^3 Hz = 1 × 10^6 Hz Given: v air = 340 m s^-1, v water = 1486 m s^-1 (a) The reflected sound The reflected wave never leaves the air — it just turns around at the surface and travels back the way it came. So it still moves at the speed of sound in air, and its wavelength is λ reflected = v air f = (340 m s^-1)/(1 × 10^6 Hz ) λ reflected = 3.4 × 10^-4 m Keeping two significant figures (matching the two figures in 340 m s ^-1 ), this stays exactly 3.4×10^-4 m — no further rounding needed. (b) The transmitted sound The transmitted wave now moves through water at 1486 m s ^-1, with the frequency unchanged at 1×10^6 Hz: λ transmitted = v water f = (1486 m s^-1)/(1 × 10^6 Hz ) λ transmitted = 1.486 × 10^-3 m Rounding once, at the end, to three significant figures: λ transmitted ≈ 1.49 × 10^-3 m Notice the transmitted wavelength is much longer than the reflected one — that's just the direct consequence of water carrying sound over four times faster than air, at the same frequency. Answer: Reflected wavelength = 3.4 × 10^-4 m; transmitted wavelength ≈ 1.49 × 10^-3 m (frequency stays at 1×10^6 Hz in both cases; only the medium's speed changes the wavelength).

Question 7

A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1? The operating frequency of the scanner is 4.2 MHz.

Accepted Answer

4.1 × 10^-4 m. Wavelength links wave speed and frequency through v = fλ, so λ = v/f. The wave equation v = fλ holds for any travelling wave, where v is the speed of the wave in the medium, f is its frequency, and λ is its wavelength. Rearranging for the wavelength gives λ = v/f Convert both quantities to SI units before substituting. The speed of sound in the tissue is v = 1.7 km s^-1 = 1.7 × 10^3 m s^-1 The operating frequency of the scanner is f = 4.2 MHz = 4.2 × 10^6 Hz A frequency this high — megahertz, not kilohertz — is exactly why ultrasonic scanning gives such a short wavelength: it is this shortness that lets it resolve small structures like a tumour, unlike audible sound. Substitute into λ = v/f. λ = (1.7 × 10^3 m s^-1)/(4.2 × 10^6 Hz ) λ = 0.404761… × 10^-3 m = 4.04761… × 10^-4 m Round to significant figures. Both given quantities, 1.7 and 4.2, carry only two significant figures, so the answer must also be quoted to two significant figures: λ ≈ 4.0 × 10^-4 m = 0.40 mm This is a submillimetre wavelength — far shorter than audible sound's wavelength (metres, for kHz sound in air) — which is precisely why ultrasound can image millimetre-scale features inside tissue: a wave can only resolve detail comparable to or larger than its own wavelength. Answer: λ ≈ 4.0 × 10^-4 m = 0.40 mm

Question 8

A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right. Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation? What are its amplitude and frequency? What is the initial phase at the origin? (d) What is the least distance between two successive crests in the wave?

Accepted Answer

(a) A travelling wave. It travels from right to left with a speed of 20 ms^-1. (b) 3.0 cm, 5.7 Hz (c) π/4 (d) 3.5 m. A travelling wave has x and t tangled together inside one sine argument; a stationary wave keeps them apart, as a product like 2A sin (kx) cos (ω t). Here y(x,t) = 3.0 sin (36t + 0.018x + π/4) has x and t added together inside a single sine, so this is a travelling wave, not a stationary one. (a) Speed and direction Match the given equation to the general travelling-wave form y = A sin (kx + ω t + φ), reading off A = 3.0 cm, k = 0.018 rad/cm, ω = 36 rad/s, φ = π/4. The sign in front of t is the part people trip over: a wave moving in the +x direction is written sin (kx - ω t), while a wave moving in the -x direction is written sin (kx + ω t) — it is the relative sign between the kx and ω t terms that fixes the direction, not the sign of k or ω on their own. Since this equation has kx and ω t with the same sign, the wave travels in the -x direction. Because positive x is defined as left to right, this wave moves from right to left. The speed of a travelling wave is v = ω/k, where ω is the angular frequency and k is the angular wave number: v = (ω)/k = (36 rad/s )/(0.018 rad/cm ) = 2.0×10^3 cm/s = 20 m/s. (b) Amplitude and frequency The amplitude is simply the coefficient of the sine, already read off above: A = 3.0 cm. The frequency comes from ω = 2π f: f = (ω)/(2π) = (36 rad/s )/(2π) = 5.7296… Hz. Rounding once, to two significant figures (matching the 36 and 0.018 in the data), f ≈ 5.7 Hz. (c) Initial phase at the origin The phase of the wave is θ(x,t) = 36t + 0.018x + π/4. "At the origin" and "initial" mean x = 0 and t = 0: θ(0,0) = 36(0) + 0.018(0) + π/4 = π/4. (d) Least distance between two successive crests Successive crests are one full spatial cycle apart, so this distance is the wavelength λ, related to the wave number by k = 2π/λ: λ = (2π)/k = (2π)/(0.018 rad/cm ) = 349.07 cm. Converting and rounding to two significant figures, λ ≈ 3.5 m. Answer: It is a travelling wave, moving in the -x direction (right to left), with speed v = 20 m/s, amplitude A = 3.0 cm, frequency f ≈ 5.7 Hz, initial phase at the origin π/4, and least distance between successive crests (wavelength) λ ≈ 3.5 m.

Question 9

For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

Accepted Answer

All the graphs are sinusoidal. They have same amplitude and frequency, but different initial phases.. A point farther along the string does not oscillate with a different size or speed — it only starts its cycle at a different moment. That time-shift is what "phase" means, and it is the only thing that changes from point to point on a travelling wave. The wave (from the previous exercise) is y(x,t) = 3.0 sin (36t + 0.018x + π/4) cm, x in cm, t in s. Here the angular frequency is ω = 36 rad s^-1, the angular wave number is k = 0.018 rad cm^-1, the amplitude is A = 3.0 cm, and φ 0=π/4 is the phase at the origin. Displacement at each of the three points. To get y versus t at a fixed x, just substitute that value of x into the wave equation — x then only adds a constant to the phase, it does not touch the t -dependence or the coefficient out front: x=0: y 1(t) = 3.0 sin (36t + π/4) cm x=2 cm: y 2(t) = 3.0 sin (36t + π/4 + (0.018)(2) ) = 3.0 sin (36t + 0.821) cm x=4 cm: y 3(t) = 3.0 sin (36t + π/4 + (0.018)(4) ) = 3.0 sin (36t + 0.857) cm (using π/4 = 0.785 rad, so the extra phases are 0.018×2=0.036 rad and 0.018×4=0.072 rad ). Shape of the graphs. Each of y 1(t), y 2(t), y 3(t) is a pure sine curve — a smooth, repeating up-and-down curve, not a straight line or a triangular wave — with: - the same amplitude, A=3.0 cm, so all three curves rise to the same peak and fall to the same trough; - the same angular frequency ω = 36 rad s^-1, giving the same period T = (2π)/(ω) = (2π)/(36 rad s^-1) ≈ 0.17 s, so all three curves repeat at exactly the same rate and would sit on top of each other if you shifted them sideways along the time axis. The only difference between the three curves is that the sine curve for x=2 cm is displaced in time relative to the one for x=0, and the curve for x=4 cm is displaced by twice as much. This is because this wave travels in the -x direction (as found in Exercise 14.8: the term 0.018x has the same sign as 36t, which is the signature of a -x -moving wave). A point farther along +x therefore feels the disturbance earlier than a point closer to the origin. Converting the extra phase to a time lead, Δ t = kx/ω: Δ t x=2 = 0.036/36 s = 1.0×10^-3 s, Δ t x=4 = 0.072/36 s = 2.0×10^-3 s. So the graph at x=2 cm is the same sine curve as at x=0, just slid 1.0 ms earlier on the time axis, and the graph at x=4 cm is slid 2.0 ms earlier still. (These shifts are tiny compared with the 0.17 s period — on an actual plot the three curves would look almost coincident, only barely staggered.) The trap here is to expect a travelling wave to look "weaker" or "slower" farther from the source, the way a damped signal might. It does not: a travelling wave (unlike a source that is dying out) carries the same amplitude and the same frequency to every point along the string. What differs from point to point is only the instant at which each point starts its cycle — i.e., only its phase. Answer: The y – t graphs at x=0, 2 cm and 4 cm are all identical sine curves of amplitude 3.0 cm and period T≈0.17 s; the curves for x=2 cm and x=4 cm are the same curve as at x=0 shifted earlier in time by 1.0×10^-3 s and 2.0×10^-3 s respectively. The oscillatory motion is identical in amplitude and frequency at every point of the string; only the phase differs from one point to another, and it is this steadily-changing phase (through the term kx ) that makes the disturbance appear to travel along the string.

Question 10

For the travelling harmonic wave y(x, t) = 2.0 cos 2π (10t - 0.0080 x + 0.35) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of 4 m, 0.5 m, λ/2, 3λ/4

Accepted Answer

(a) 6.4 π rad (b) 0.8 π rad (c) π rad (d) (π/2) rad. At one instant, the phase difference between two points on a travelling wave depends only on how far apart they are: Δφ = k Δ x, where k is the angular wave number. Start by putting the wave in the standard form y = a cos (ω t - kx + φ 0), in which a is the amplitude, ω the angular frequency, k the angular wave number and φ 0 the initial phase constant. Multiplying the 2π through the bracket, y(x,t) = 2.0 cos [ 2π(10t - 0.0080x + 0.35) ] = 2.0 cos (20π t - 0.016π x + 0.70π ) cm Reading off the coefficients: a = 2.0 cm, ω = 20π rad s^-1, k = 0.016π rad cm^-1, φ 0 = 0.70π rad The wavelength follows from k = 2π/λ: λ = (2π)/k = (2π rad )/(0.016π rad cm^-1) = 125 cm = 1.25 m Watch the units here. The problem states that x is in centimetres, so k comes out in radians per centimetre. Every separation must be converted to centimetres before it is multiplied by k; using metres makes the answer 100 times too small. This is the single step most people get wrong in this question. Now the phase difference itself. The phase of the point at x is φ(x) = ω t - kx + φ 0. For two points x 1 and x 2 = x 1 + Δ x at the same instant t, Δφ = φ(x 2) - φ(x 1) = k Δ x = (2π)/(λ) Δ x The ω t term and the constant φ 0 are the same for both points, so they cancel — which is why the answer does not depend on when you look. (a) Separation 4 m. Convert: Δ x = 4 m = 400 cm. Δφ = (0.016π rad cm^-1)(400 cm ) = 6.4π rad ≈ 20 rad (b) Separation 0.5 m. Convert: Δ x = 0.5 m = 50 cm. Δφ = (0.016π rad cm^-1)(50 cm ) = 0.8π rad ≈ 2.5 rad (c) Separation λ/2. Here the separation is given as a fraction of the wavelength, so λ cancels and no number needs substituting: Δφ = (2π)/(λ)·(λ)/2 = π rad ≈ 3.14 rad Two points half a wavelength apart are exactly out of phase: when one is at a crest the other is at a trough. (d) Separation 3λ/4. Δφ = (2π)/(λ)·(3λ)/4 = (3π)/2 rad ≈ 4.71 rad Because cos repeats every 2π, a lag of 3π/2 is the same state of motion as a lead of π/2: (3π)/2 - 2π = -(π)/2 So the two points can equally be described as π/2 rad out of step, with the far point leading rather than lagging. NCERT prints π/2 rad for this part, having reduced the angle to the smallest equivalent one. That reduction is legitimate — but note that the book does not apply it to parts (a) and (b), where 6.4π would have collapsed to 0.4π by the same rule. Taken straight from Δφ = k Δ x, and consistently with the first three parts, the value is 3π/2 rad; quote π/2 rad only if you are deliberately asking for the smallest equivalent phase difference. Neither number describes a different physical situation. On significant figures: the coefficient 0.0080 carries two significant figures, so the decimal values in (a) and (b) round to 20 rad and 2.5 rad. Parts (c) and (d) are exact multiples of π — the separations were given as fractions of λ, so k cancels and its precision never enters. Answer: Δφ = k Δ x with k = 0.016π rad cm^-1 and λ = 125 cm = 1.25 m; (a) 6.4π rad ≈ 20 rad; (b) 0.8π rad ≈ 2.5 rad; (c) π rad ≈ 3.14 rad; (d) 3π/2 rad ≈ 4.71 rad, equivalently π/2 rad if the angle is reduced modulo 2π, which is the form NCERT prints.

Question 11

The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin 2 3 π x ⎛ ⎝⎜ ⎞ ⎠⎟ cos (120 πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10^-2 kg. Answer the following: Does the function represent a travelling wave or a stationary wave? Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave? (c) Determine the tension in the string.

Accepted Answer

(a) Stationary wave (b) l = 3 m, n = 60 Hz, and v = 180 m s^-1 for each wave (c ) 648 N. A stationary wave is a product of a position-only function and a time-only function; a travelling wave must depend only on the single combination x - vt (or x+vt ). The given displacement is y(x,t) = 0.06 sin ((2π x)/3) cos (120π t) Here the x-dependence and the t-dependence sit in two separate factors — sin (2π x/3) fixes a shape along the string, and cos (120π t) makes the amplitude of that fixed shape rise and fall in time. A travelling wave, by contrast, must be expressible as a function of x and t only through the combination x-vt, so that the whole waveform slides along without changing shape. Because y(x,t) here cannot be written that way, it represents a stationary (standing) wave, not a travelling one. (Every point on the string still oscillates in time — that is not the test. The test is whether the pattern itself moves along x, and here it does not: the points where sin (2π x/3)=0 stay at rest for all t — these are nodes.) A standing wave is what you get when two identical waves travelling in opposite directions are superposed — the product-to-sum identity makes that split explicit. Use sin A cos B = 12 [ sin (A+B)+ sin (A-B) ] with A = (2π x)/3, B = 120π t: y(x,t) = 0.06 sin ((2π x)/3) cos (120π t) = 0.03 sin ((2π x)/3 - 120π t) + 0.03 sin ((2π x)/3 + 120π t) A term of the form sin (kx-ω t) travels in the +x direction (to keep the argument fixed as t grows, x must grow too); a term of the form sin (kx+ω t) travels in the -x direction. So the standing wave is the sum of y 1 = 0.03 sin ((2π x)/3 - 120π t) m, travelling in +x, y 2 = 0.03 sin ((2π x)/3 + 120π t) m, travelling in -x, each of amplitude 0.03 m (half the standing-wave amplitude — this halving is exactly what superposition of two 0.03 m waves reconstructs). Reading off the wave number and angular frequency common to both: k = (2π)/3 rad/m, ω = 120π rad/s. These come straight from the given equation, so they carry no rounding error. Wavelength: k = (2π)/(λ) ⇒ λ = (2π)/k = (2π)/(2π/3) = 3 m Frequency: ω = 2π f ⇒ f = (ω)/(2π) = (120π)/(2π) = 60 Hz Speed: v = fλ = 60 Hz × 3 m = 180 m/s Both component waves — same λ, same f, same speed, opposite direction — have λ = 3 m, f = 60 Hz, v = 180 m/s. (c) The wave speed on a stretched string ties directly to its tension through v=√ T/μ, where μ is mass per unit length — not the total mass. μ = ( mass )/( length ) = (3.0×10^-2 kg )/(1.5 m ) = 2.0×10^-2 kg/m (A common slip here is plugging the string's total mass into v=√ T/m directly — the formula needs mass per metre, so the division by length is not optional.) From v = √ T/μ, squaring gives T = μ v^2: T = (2.0×10^-2 kg/m )(180 m/s )^2 = (2.0×10^-2 kg/m )(3.24×10^4 m^2/ s^2) = 648 kg·m/s^2 = 648 N The mass (3.0×10⁻² kg) and length (1.5 m) each carry only 2 significant figures, so μ — and hence T — is good to 2 significant figures: rounding once, at the end, T ≈ 6.5×10^2 N. Answer: (a) A stationary (standing) wave, since x and t appear in separate factors, not as x-vt. (b) y = 0.03 sin (2π x/3 - 120π t) + 0.03 sin (2π x/3 + 120π t) m, i.e. two waves of amplitude 0.03 m travelling in opposite directions (+x and −x), each with λ = 3 m, f = 60 Hz, v = 180 m/s. (c) T = 648 N ≈ 6.5×10^2 N (2 significant figures).

Question 12

(i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

Accepted Answer

(a) All the points except the nodes on the string have the same frequency and phase, but not the same amplitude. (b) 0.042 m. In a stationary wave the time factor is shared by every point, so the only thing that can differ from point to point is the size of the swing — and the sign of that swing. The string is described by y(x,t) = 0.06 sin ((2π x)/3) cos (120π t) m, 0 ≤ x ≤ 1.5 m. This has the standard stationary-wave form y = A(x) cos (ω t), where - A(x) = 0.06 sin ((2π x)/3) m is the amplitude of the point at position x, - ω = 120π rad s^-1 is the angular frequency, - k = (2π)/3 rad m^-1 is the wave number. Every point on the string obeys y = A(x) cos (ω t) — the same function of time, only scaled by A(x). That single observation answers all three parts of (i). (i)(a) Frequency — yes, the same for all points. The time dependence is cos (ω t) with ω = 120π rad s^-1, and ω = 2πν, so ν = (ω)/(2π) = 120π rad s^-1 2π = 60 Hz. Nothing in cos (120π t) depends on x, so every oscillating point of the string completes its cycle in the same time: all points vibrate at 60 Hz. (The nodes are the exception only in the trivial sense that they never move at all, so no frequency can be assigned to them.) (i)(b) Phase — yes, all points except the nodes are in phase. You must check where the nodes are before saying this. The phase of a point is decided by the sign of A(x): if A(x) 0 the point moves with cos (ω t); if A(x) < 0 it moves with - cos (ω t), i.e. exactly π out of phase. So the question becomes: does sin (2π x/3) change sign anywhere on this string? Nodes are the points where A(x) = 0: sin ((2π x)/3) = 0 ⇒ (2π x)/3 = nπ ⇒ x = 1.5 n m, n = 0, 1, 2, … Inside the string 0 ≤ x ≤ 1.5 m, that gives nodes at x = 0 and x = 1.5 m — the two clamped ends — and nowhere else. Equivalently, λ = 2π/k = 3 m, and L = 1.5 m = λ/2: the string is vibrating in its fundamental mode, one single loop. With no interior node, sin (2π x/3) 0 for every 0 < x < 1.5 m. Every point therefore carries the same + cos (120π t) time dependence: they all reach maximum displacement together, all pass through zero together, and all move to the same side of the string at the same instant. All points except the nodes oscillate in phase. The step people get wrong: it is true in general that adjacent loops of a stationary wave are π out of phase — a point in one loop is going up while a point in the next loop is going down. That is a real effect, but it needs at least two loops, i.e. at least one node between the ends. This string has none, so the rule never gets a chance to bite here. Always locate the nodes before deciding the phase; do not import the multi-loop picture into a fundamental. (i)(c) Amplitude — no, it is different at every point. The amplitude is A(x) = 0.06 sin (2π x/3) m, which plainly depends on x: it is 0 at the nodes x = 0 and x = 1.5 m, and rises to its largest value 0.06 m at the antinode in the middle, x = 0.75 m, where sin (2π x/3) = sin (π/2) = 1. So the amplitude is not the same for all points. (ii) Amplitude of the point 0.375 m from one end. Measure x from the clamped end at x = 0, so x = 0.375 m, and substitute into A(x): A(0.375) = 0.06 × sin ((2π × 0.375)/3) m = 0.06 × sin ((0.75π)/3) m = 0.06 × sin ((π)/4) m. The angle here is in radians, not degrees — kx is always a radian measure, and π/4 rad = 45^°. Using sin (π/4) = 1/√ 2: A(0.375) = 0.06/(√ 2 ) m = 0.042426… m. Rounding once, at the end, to the two significant figures carried by the data: A(0.375 m ) = 0.042 m = 4.2 cm. A sanity check on the size: x = 0.375 m is halfway between the node at x = 0 and the antinode at x = 0.75 m, so the amplitude should be a good fraction of the antinode's 0.06 m but clearly below it — and 0.042 m is 1/√ 2 of it, which is exactly what sin 45^° demands. Answer: (i)(a) Yes — every oscillating point has the same frequency, ν = 60 Hz. (b) Yes — the string is vibrating in one loop ( λ = 3 m, L = λ/2 ), with nodes only at the two clamped ends, so 0.06 sin (2π x/3) never changes sign and all points except the nodes oscillate in phase. (c) No — the amplitude varies with position as A(x) = 0.06 sin (2π x/3) m, from 0 at the nodes to 0.06 m at the antinode x = 0.75 m. (ii) A = 0.06/√ 2 = 0.042 m (4.2 cm), to two significant figures.

Question 13

Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all: (a) y = 2 cos (3x) sin (10t) (b) y x vt = − 2 y = 3 sin (5x - 0.5t) + 4 cos (5x - 0.5t) (d) y = cos x sin t + cos 2x sin 2t

Accepted Answer

(a) Stationary wave. (b) Unacceptable function for any wave. (c) Travelling harmonic wave. (d) Superposition of two stationary waves.. Two tests decide every part — but only after a third one is passed. A function represents a travelling wave if x and t enter only through the single combination x-vt or x+vt; it represents a stationary wave if it separates as (a shape in x ) × (an oscillation in t ), so that the pattern stays put and only breathes up and down. Before either test, the function must be a physically possible displacement at all: real, finite and single-valued for every x and every t, and it must satisfy the wave equation (∂^2y)/(∂ x^2)=1/(v^2) (∂^2y)/(∂ t^2) where v is the wave speed. A function that fails that is the "none at all" case. (a) y = 2 cos (3x) sin (10t) This is already a product of a function of x alone and a function of t alone, which is the signature of a stationary wave. Comparing with y = 2a cos kx sin ω t: k = 3, ω = 10. Every particle performs simple harmonic motion of the same angular frequency ω = 10, with an amplitude 2 cos 3x that depends on where the particle sits. The nodes are the points that never move, cos 3x = 0: 3x = (2n+1)(π)/2 ⇒ x = (2n+1)(π)/6, n = 0,1,2,… Using 2 sin A cos B = sin (A+B)+ sin (A-B) with A = 10t, B = 3x, y = sin (10t+3x) + sin (10t-3x), so it is the sum of two identical harmonic waves running in opposite directions with speed v = ω/k = 10/3 ≈ 3.3 units. That is exactly how a stationary wave is made. Stationary wave. (b) y = 2√ x-vt This is the part that traps people: x and t do appear only through x-vt, so the travelling-wave form is satisfied and it is tempting to tick "travelling". The form is necessary but not sufficient — the function must also be a possible displacement everywhere. - For x < vt the quantity x-vt is negative and y is imaginary. A displacement cannot be imaginary. - As x-vt → ∞, y → ∞. A displacement cannot be unbounded; no finite energy source could produce it. So there is no region of x and t in which this is a legitimate wave. None at all — an unacceptable function for any wave. (c) y = 3 sin (5x-0.5t) + 4 cos (5x-0.5t) Write θ = 5x - 0.5t and combine the two terms using a sin θ + b cos θ = R sin (θ+φ) with R = √ a^2+b^2 and tan φ = b/a: R = √ 3^2+4^2 = √ 9+16 = √ 25 = 5, φ = tan^-1 (4/3) = 0.927 rad (53.1^°). y = 5 sin (5x - 0.5t + 0.927). It is a single sine of one argument, and that argument contains x and t only through 5x - 0.5t = 5(x - 0.1 t), which is the combination x - vt. Comparing with y = A sin (kx-ω t+φ): k = 5, ω = 0.5, v = (ω)/k = 0.5/5 = 0.1. The minus sign between kx and ω t means the pattern moves towards increasing x — a direction is part of this answer, not an optional extra. Amplitude 5, speed 0.1 in the units of the question ( 0.1 m s^-1 if x is in metres and t in seconds), quoted to the one significant figure the data carry. Travelling harmonic wave, moving in the +x direction. (d) y = cos x sin t + cos 2x sin 2t Take each term on its own. cos x sin t is separable, so it is a stationary wave with k 1 = 1, ω 1 = 1; cos 2x sin 2t is separable, so it is a stationary wave with k 2 = 2, ω 2 = 2. Both have the same speed ω/k = 1, so both are legitimate waves on the same medium, and the sum of two solutions of the wave equation is again a solution (superposition principle). Checking directly for the whole function: (∂^2y)/(∂ x^2) = - cos x sin t - 4 cos 2x sin 2t, (∂^2y)/(∂ t^2) = - cos x sin t - 4 cos 2x sin 2t, so ∂^2y/∂ x^2 = ∂^2y/∂ t^2, which is the wave equation with v = 1. It is also bounded, y ≤ 2. So this is an acceptable wave function; it cannot be put in the "none at all" box, and that is the box people wrongly reach for here. Category (iii) is for functions like (b) that could never be a displacement — not for functions that merely look complicated. Is it travelling? At t = 0, y = cos x sin 0 + cos 2x sin 0 = 0 for every x. A travelling wave y = f(x-vt) just slides a fixed profile along, so if the profile is zero everywhere at one instant, f is identically zero and the wave is zero for all time. Here y is not zero at later times — at x = π/2, for instance, y = - sin 2t. So it does not travel, and no energy is carried along the string on average. Not travelling, and a genuine wave, leaves the stationary category. The honest description is the one to write down: it is the superposition of two stationary waves, of angular frequencies 1 and 2. It is worth seeing why that phrasing is used rather than "a stationary wave" flat: the sum does not factorise as (function of x ) × (function of t ), and it has no permanent nodes — a node would need cos x = 0 and cos 2x = 0 at the same x, but cos 2x = 2 cos^2x - 1 = -1 wherever cos x = 0. Each particle still oscillates about its mean position with a fixed period 2π, but with two frequencies mixed in rather than one. The pattern stays where it is; it does not propagate. Stationary — a superposition of two stationary waves. Answer: (a) stationary wave (nodes at x=(2n+1)π/6 ); (b) none at all — 2√ x-vt is imaginary for x<vt and unbounded for large x-vt, so it is an unacceptable function for any wave; (c) travelling harmonic wave, y=5 sin (5x-0.5t+0.927), moving in the +x direction with speed v=ω/k=0.1 ( 0.1 m s^-1 for x in metres, t in seconds); (d) stationary — the superposition of two stationary waves of angular frequencies 1 and 2.

Question 14

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10^-2 kg and its linear mass density is 4.0 × 10^-2 kg m^-1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Accepted Answer

(a) 79 m s^-1 (b) 248 N. A string fixed at both ends vibrating in its fundamental mode fits exactly half a wavelength between the supports — that single fact converts the wire's physical length into its wavelength. Step 1 — Find the length of the wire. Linear mass density is mass per unit length, μ = M/L, so the length is L = M/(μ) With M = 3.5 × 10^-2 kg and μ = 4.0 × 10^-2 kg m^-1: L = (3.5 × 10^-2 kg )/(4.0 × 10^-2 kg m^-1) = 0.875 m Step 2 — Speed of the transverse wave, from the fundamental-mode condition. For a string fixed at both ends, the fundamental (first-harmonic) standing wave has a node at each support and one antinode in the middle — the wire spans half a wavelength: L = (λ)/2 ⇒ λ = 2L The wave speed is related to frequency and wavelength by v = fλ, so v = f(2L) = 2Lf Substituting L = 0.875 m and f = 45 Hz: v = 2 × 0.875 m × 45 Hz = 78.75 m s^-1 Keep this unrounded value for the next step — rounding now and squaring later would compound the error. Rounded only for reporting (two significant figures, matching the two-sig-fig data given): v ≈ 79 m s^-1. Step 3 — Tension in the string, from the wave-speed formula. The speed of a transverse wave on a stretched string is v = √ T/μ, where T is the tension and μ the linear mass density. Squaring and rearranging: T = μ v^2 Using the unrounded speed v = 78.75 m s^-1: T = 4.0 × 10^-2 kg m^-1 × (78.75 m s^-1)^2 = 4.0 × 10^-2 × 6201.5625 N = 248.06 N Every input in this problem (45 Hz, 3.5×10^-2 kg, 4.0×10^-2 kg m ^-1 ) carries only two significant figures, so the final answers should not claim more precision than that. Rounding once, at the very end: T ≈ 2.5 × 10^2 N A common slip here is mixing up mass and linear mass density, or using the rounded 79 m/s (instead of the exact 78.75 m/s) inside T = μ v^2 — small rounding errors get squared and can shift the tension by several newtons. Answer: (a) speed of the transverse wave v ≈ 79 m s^-1; (b) tension T ≈ 2.5 × 10^2 N (≈ 250 N)

Question 15

A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Accepted Answer

347 m s^-1 Hint: v n = l v) n ( 4 1 2 −; n = 1,2,3,….for a pipe with one end closed. A tube with a piston at one end is a pipe closed at one end, and such a pipe resonates when its length is an odd number of quarter-wavelengths — so the first resonance measures λ/4 directly. The piston face is a rigid wall, so the air there cannot move: that end is a displacement node. The open end is a displacement antinode. Fitting a standing wave between a node and an antinode requires l n=((2n-1)λ)/4, n=1,2,3,… which is the same statement as the hint v n = (4 l ν)/2n-1. Here l is the length of air column, λ the wavelength of sound in the air, ν the frequency and v the speed of sound. An aside on what is fixed and what is not: the frequency is set by the tuning fork, ν = 340 Hz, and never changes. Sliding the piston does not change the pitch — it changes the length of the air column until one of the column's natural frequencies matches the fork. So ν is an input here, not something the tube decides. The first resonance. The shortest resonating length is the fundamental, n=1: l 1=(λ)/4 ⇒ λ = 4l 1 = 4× 0.255 m = 1.02 m The wave relation v=νλ then gives v = 340 s^-1× 1.02 m = 346.8 m s^-1 Rounding once, at the end, to the three significant figures carried by both 340 Hz and 25.5 cm: v ≈ 347 m s^-1 What the second length is. With λ = 1.02 m, the next resonance is n=2: l 2=(3λ)/4=(3× 1.02 m )/4=0.765 m =76.5 cm and the one after that would be 5λ/4 = 127.5 cm, longer than the metre-long tube — which is why the experiment finds exactly two resonances and no more. So the second reading is identified as the first overtone (third harmonic). Note that it is 79.3 cm as measured, about 2.8 cm longer than the ideal 76.5 cm. That gap is the "edge effect" the question tells us to neglect: at a real open end the antinode sits a little outside the tube, so a measured column is not exactly a quarter-wavelength. Having been told to neglect it, we take the clean quarter-wave relation and read the speed off the fundamental. The step people get wrong here. It is tempting to use the spacing of successive resonances, since consecutive resonances of any pipe are half a wavelength apart: λ = 2(l 2-l 1)=2(0.793-0.255) m =1.076 m, v=340×1.076=366 m s^-1 That method is the right one when you are allowing for an end correction, because the correction cancels in the difference. It is the wrong one here, for two reasons. First, the problem states that edge effects are negligible; once that is assumed, both λ=4l 1 and λ=2(l 2-l 1) would have to hold, and they cannot both be true of these numbers — the data are over-determined and disagree by about 5 %. Second, and decisively, check the two candidates against the temperature. In air, v ≈ (331.3 + 0.606 T) m s^-1 with T in ^° C: 347 m s^-1⇒ T≈ 26^° C, 366 m s^-1⇒ T≈ 57^° C A room at 26^° C is an ordinary laboratory; 57^° C is not a temperature anyone runs a resonance-tube experiment at. The fundamental reading gives the physically sensible answer, and the 2.8 cm surplus in the second reading is a blemish in the quoted data rather than a signal to switch methods. Answer: v = νλ = 340 Hz × 4(0.255 m ) = 346.8 ≈ 347 m s^-1 (three significant figures), corresponding to an air temperature of about 26^° C.

Question 16

A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?

Accepted Answer

5.06 km s^-1. A clamp fixes the rod, so the clamped point cannot move — it is a displacement node, not an antinode. The two free ends, since nothing holds them, vibrate with maximum amplitude — they are antinodes. Getting this boundary condition right is the whole problem; nothing else needs a picture. Setting up the standing wave For a standing wave, the distance between a node and the nearest antinode is always a quarter wavelength: node-to-antinode distance = (λ)/4 The rod has length L = 100 cm = 1.00 m, clamped exactly at its middle. So the node sits at the centre, and each free end is an antinode a distance l = L/2 = (1.00 m )/2 = 0.50 m away from that node. This half-length is what spans the quarter wavelength — not the full rod. (A common slip here is to set L = λ or L = λ/2, which is the condition for a rod free at both ends or fixed at both ends. A rod clamped only in the middle is a different boundary condition, with the node in the centre and antinodes at the two free ends.) Finding the wavelength (λ)/4 = l = 0.50 m λ = 4 × 0.50 m = 2.00 m Finding the speed The wave equation relates speed, frequency and wavelength: v = fλ where f is the fundamental frequency of longitudinal vibration, given as 2.53 kHz = 2.53 × 10^3 Hz. Substituting: v = (2.53 × 10^3 Hz ) × (2.00 m ) v = 5.06 × 10^3 m/s Both the given frequency (2.53 kHz) and the given length (100 cm, i.e. 1.00 m) carry three significant figures, so the answer is kept to three significant figures: 5.06 × 10^3 m/s, i.e. 5.06 km/s. This is speed of sound (longitudinal wave speed) in the steel of the rod, not in air — steel is far stiffer than air, which is exactly why the value comes out thousands of times larger than the ~340 m/s speed of sound in air. Answer: The speed of sound (longitudinal wave) in steel is 5.06 × 10^3 m/s = 5.06 km/s.

Question 17

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s^-1 ).

Accepted Answer

First harmonic (fundamental); No.. A pipe closed at one end supports only the odd harmonics — find its fundamental frequency first, then see which multiple lands near 430 Hz. For an air column closed at one end, one end must be a displacement node and the open end an antinode. This fixes the allowed wavelengths to L = (2n-1)(λ n)/4, n = 1, 2, 3, … so the resonant frequencies are f n = (2n-1)v/4L, n = 1, 2, 3, … Only the odd multiples of the fundamental appear (1st, 3rd, 5th harmonic, …) — a closed pipe never sounds its even harmonics. Here the pipe length is L = 20 cm = 0.20 m and the speed of sound is v = 340 m s^-1. Fundamental (first harmonic), n=1: f 1 = v/4L = (340 m s^-1)/(4 × 0.20 m ) = 340/0.80 Hz = 425 Hz Third harmonic, n=2: f 3 = 3f 1 = 3 × 425 Hz = 1275 Hz The source frequency is 430 Hz. Compare it with the ladder of allowed frequencies: 425 Hz, 1275 Hz, 2125 Hz, … The third harmonic (1275 Hz) and every harmonic above it are nowhere near 430 Hz, but the fundamental, 425 Hz, sits only 5 Hz away from the driving frequency — a difference of about 1%, well inside the rounding already present in the given data ( L and v are given to only two significant figures). So the 430 Hz source drives the pipe at its fundamental: the first harmonic. Now check the same pipe with both ends open. An open-open pipe has an antinode at each end, so it supports every integer harmonic: f n' = nv/2L, n = 1, 2, 3, … Its fundamental is f 1' = v/2L = (340 m s^-1)/(2 × 0.20 m ) = 340/0.40 Hz = 850 Hz and its allowed frequencies are 850 Hz, 1700 Hz, 2550 Hz, … A 430 Hz source is close to none of these — it is roughly half of 850 Hz, but an open-open pipe has no mode at a half-integer multiple of its fundamental (only whole multiples occur, unlike the closed pipe's odd-multiple ladder built on a smaller base frequency). So the 430 Hz source would not be in resonance with the same pipe if both ends were open. The reason the two cases differ so sharply: opening the second end raises the fundamental from v/4L to v/2L — it doubles — while also switching the harmonic content from odd multiples only to all integer multiples. Neither ladder happens to pass near 430 Hz once that end is opened. Answer: The 430 Hz source resonantly excites the first harmonic (fundamental, f 1 = 425 Hz ) of the pipe closed at one end. With both ends open the fundamental becomes 850 Hz (and all higher harmonics are integer multiples of it), so the 430 Hz source would not be in resonance with the pipe in that case.

Question 18

Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Accepted Answer

318 Hz BIBLIOGRAPHY TEXTBOOKS For additional reading on the topics covered in this book, you may like to consult one or more of the following books. Some of these books however are more advanced and contain many more topics than this book. 1. Ordinary Level Physics, A.F. Abbott, Arnold-Heinemann (1984). 2. Advanced Level Physics, M. Nelkon and P. Parker, 6^th Edition Arnold- Heinemann (1987). 3. Advanced Physics, Tom Duncan, John Murray (2000). 4. Fundamentals of Physics, David Halliday, Robert Resnick and Jearl Walker, 7th Edition John Wily (2004). 5. University Physics, H.D. Young, M.W. Zemansky and F.W. Sears, Narosa Pub. House (1982). 6. Problems in Elementary Physics, B. Bukhovtsa, V. Krivchenkov, G. Myakishev and V. Shalnov, MIR Publishers, (1971). 7. Lectures on Physics (3 volumes), R.P. Feynman, Addision - Wesley (1965). 8. Berkeley Physics Course (5 volumes) McGraw Hill (1965). a. Vol. 1 - Mechanics: (Kittel, Knight and Ruderman) b. Vol. 2 - Electricity and Magnetism (E.M. Purcell) c. Vol. 3 - Waves and Oscillations (Frank S. Craw-ford) d. Vol. 4 - Quantum Physics (Wichmann) e. Vol. 5 - Statistical Physics (F. Reif) 9. Fundamental University Physics, M. Alonso and E. J. Finn, Addison - Wesley (1967). 10. College Physics, R.L. Weber, K.V. Manning, M.W. White and G.A. Weygand, Tata McGraw Hill (1977). 11. Physics: Foundations and Frontiers, G. Gamow and J.M. Cleveland, Tata McGraw Hill (1978). 12. Physics for the Inquiring Mind, E.M. Rogers, Princeton University Press (1960) 13. PSSC Physics Course, DC Heath and Co. (1965) Indian Edition, NCERT (1967) 14. Physics Advanced Level, Jim Breithampt, Stanley Thornes Publishers (2000). 15. Physics, Patrick Fullick, Heinemann (2000). 16. Conceptual Physics, Paul G. Hewitt, Addision-Wesley (1998). 17. College Physics, Raymond A. Serway and Jerry S. Faughn, Harcourt Brace and Co. (1999). 18. University Physics, Harris Benson, John Wiley (1996). 19. University Physics, William P. Crummet and Arthur B. Western, Wm.C. Brown (1994). 20. General Physics, Morton M. Sternheim and Joseph W. Kane, John Wiley (1988). 21. Physics, Hans C. Ohanian, W.W. Norton (1989). 22. Advanced Physics, Keith Gibbs, Cambridge University Press(1996). 23. Understanding Basic Mechanics, F. Reif, John Wiley (1995). 24. College Physics, Jerry D. Wilson and Anthony J. Buffa, Prentice-Hall (1997). 25. Senior Physics, Part - I, I.K. Kikoin and A.K. Kikoin, Mir Publishers (1987). 26. Senior Physics, Part - II, B. Bekhovtsev, Mir Publishers (1988). 27. Understanding Physics, K. Cummings, Patrick J. Cooney, Priscilla W. Laws and Edward F. Redish, John Wiley (2005) 28. Essentials of Physics, John D. Cutnell and Kenneth W. Johnson, John Wiley (2005) GENERAL BOOKS For instructive and entertaining general reading on science, you may like to read some of the following books. Remember however, that many of these books are written at a level far beyond the level of the present book. 1. Mr. Tompkins in paperback, G. Gamow, Cambridge University Press (1967). 2. The Universe and Dr. Einstein, C. Barnett, Time Inc. New York (1962). 3. Thirty years that Shook Physics, G. Gamow, Double Day, New York (1966). 4. Surely You’re Joking, Mr. Feynman, R.P. Feynman, Bantam books (1986). 5. One, Two, Three… Infinity, G. Gamow, Viking Inc. (1961). 6. The Meaning of Relativity, A. Einstein, (Indian Edition) Oxford and IBH Pub. Co (1965). 7. Atomic Theory and the Description of Nature, Niels Bohr, Cambridge (1934). 8. The Physical Principles of Quantum Theory, W. Heisenberg, University of Chicago Press (1930). 9. The Physics- Astronomy Frontier, F. Hoyle and J.V. Narlikar, W.H. Freeman (1980). 10. The Flying Circus of Physics with Answer, J. Walker, John Wiley and Sons (1977). 11. Physics for Everyone (series), L.D. Landau and A.I. Kitaigorodski, MIR Publisher (1978). Book 1: Physical Bodies Book 2: Molecules Book 3: Electrons Book 4: Photons and Nuclei. 12. Physics can be Fun, Y. Perelman, MIR Publishers (1986). 13. Power of Ten, Philip Morrison and Eames, W.H. Freeman (1985). 14. Physics in your Kitchen Lab., I.K. Kikoin, MIR Publishers (1985). 15. How Things Work: The Physics of Everyday Life, Louis A. Bloomfield, John Wiley (2005) 16. Physics Matters: An Introduction to Conceptual Physics, James Trefil and Robert M. Hazen, John Wiley (2004). PHYSICS 308 A Absolute scale temperature 280 Absolute zero 280 Acceleration (linear) 45 Acceleration due to gravity 49,189 Accuracy 22 Action-reaction 97 Addition of vectors 67 Adiabatic process 311, 312 Aerofoil 262 Air resistance 79 Amplitude 344, 372 Angle of contact 267, 268 Angstrom 21 Angular Acceleration 154 Angular displacement 342 Angular frequency 344, 373 Angular momentum 155 Angular velocity 152 Angular wave number 372 Antinodes 381,382 Archimedes Principle 255 Area expansion 281 Atmospheric pressure 253 Average acceleration 45, 74 Average speed 42 Average velocity 42 Avogardo's law 325 B Banked road 104 Barometer 254 Beat frequency 383 Beats 382, 383 Bending of beam 244 Bernoulli's Principle 258 Blood pressure 276 Boiling point 287 Boyle's law 326 Buckling 244 INDEX Bulk modulus 242 Buoyant force 255 C Calorimeter 285 Capillary rise 268 Capillary waves 370 Carnot engine 316 Central forces 186 Centre of Gravity 161 Centre of mass 144 Centripetal acceleration 81 Centripetal force 104 Change of state 287 Charle's law 326 Chemical Energy 126 Circular motion 104 Clausius statement 315 Coefficient of area expansion 283 Coefficient of linear expansion 281 Coefficient of performance 314 Coefficient of static friction 101 Coefficient of viscosity 262 Coefficient of volume expansion 281 Cold reservoir 313 Collision 129 Collision in two dimensions 131 Compressibility 242, 243 Compressions 368, 369, 374 Compressive stress 236, 243 Conduction 290 Conservation laws 12 Conservation of angular momentum 157, 173 Conservation of Mechanical Energy 121 Conservation of momentum 98 Conservative force 121 Constant acceleration 46,75 Contact force 100 Convection 293 Couple 159 Crest 371 Cyclic process 312 D Dalton's law of partial pressure 325 Damped oscillations 355 Damped simple Harmonic motion 355 Damping constant 355 Damping force 355 Derived units 16 Detergent action 269 Diastolic pressure 277 Differential calculus 61 Dimensional analysis 32 Dimensions 31 Displacement vector 66 Displacement 40 Doppler effect 385, 386 Doppler shift 387 Driving frequency 358 Dynamics of rotational motion 169 E Efficiency of heat engine 313 Elastic Collision 129 Elastic deformation 236, 238 Elastic limit 238 Elastic moduli 239 Elasticity 235 Elastomers 239 Electromagnetic force 8 Energy 117 Equality of vectors 66 Equation of continuity 257 Equilibrium of a particle 99 Equilibrium of Rigid body 158 Equilibrium position 341, 342, 353 Errors in measurement 22 Escape speed 193 F First law of Thermodynamics 307 Fluid pressure 251 Force 94 Forced frequency 357 Forced oscillations 357, 358 Fracture point 238 Free Fall 49 Free-body diagram 100 Frequency of periodic motion 342, 372 Friction 101 Fundamental Forces 6 Fundamental mode 381 Fusion 287 G Gauge pressure 253 Geocentric model 183 Geostationary satellite 196 Gravitational constant 189 Gravitational Force 8, 192 Gravitational potential energy 191 Gravity waves 370 H Harmonic frequency 380, 381 Harmonics 380, 381 Heat capacity 284 Heat engines 313 Heat pumps 313 Heat 279 Heliocentric model 183 Hertz 343 Hooke's law 238 Horizontal range 78 Hot reservoir 313 Hydraulic brakes 255, 256 Hydraulic lift 255, 256 Hydraulic machines 255 Hydraulic pressure 238 Hydraulic stress 238, 243 Hydrostatic paradox 253 I Ideal gas equation 280 Ideal gas 280, 325 Impulse 96 Inelastic collision 129 Initial phase angle 372 Instantaneous acceleration 74 Instantaneous speed 45 Instantaneous velocity 43 Interference 377 Internal energy 306, 330 Irreversible engine 315, 317 Irreversible processes 315 Isobaric process 311, 312 Isochoric process 311, 312 Isotherm 310 Isothermal process 311 K Kelvin-Planck statement 315 Kepler's laws of planetary motion 184 Kinematics of Rotational Motion 167 Kinematics 39 Kinetic energy of rolling motion 174 Kinetic Energy 117 Kinetic interpretation of temperature 329 Kinetic theory of gases 328 L Laminar flow 258, 264 Laplace correction 376 310 PHYSICS Latent heat of fusion 290 Latent heat of vaporisation 290 Latent heat 289 Law of cosine 72 Law of equipartition of energy 332 Law of Inertia 90 Law of sine 72 Linear expansion 281 Linear harmonic oscillator 349, 351 Linear momentum 155 Longitudinal strain 236 Longitudinal strain 236, 239 Longitudinal stress 236 Longitudinal Wave 369, 376 M Magnus effect 261 Manometer 254 Mass Energy Equivalence 126 Maximum height of projectile 78 Maxwell Distribution 331 Mean free path 324, 335 Measurement of length 18 Measurement of mass 21 Measurement of temperature 279 Measurement of time 22 Melting point 286 Modes 380 Modulus of elasticity 238 Modulus of rigidity 242 Molar specific heat capacity 284, 308 at constant pressure Molar specific heat capacity 284, 308 at constant volume Molar specific heat capacity 284 Molecular nature of matter 323 Moment of Inertia 163 Momentum 93 Motion in a plane 72 Multiplication of vectors 67 Musical instruments 384 N Natural frequency 358 Newton's first law of motion 91 Newton's Law of cooling 295 Newton's law of gravitation 185 Newton's second law of motion 93 Newton's third law of motion 96 Newtons' formula for speed of sound 377 Nodes 381 Normal Modes 381, 382, 384 Note 384, 385 Nuclear Energy 126 Null vector 68 O Odd harmonics 382 Orbital velocity/speed 194 Order of magnitude 28 Oscillations 342 Oscillatory motion 342 P Parallax method 18 Parallelogram law of addition of vectors 66 Pascal's law 252 Path length 40 Path of projectile 78 Periodic force 358 Periodic motion 342 Periodic time 342 Permanent set 238 Phase angle 344 Phase constant 344 Pipe open at both ends 382 Pipe open at one end 381 Pitch 384 Plastic deformation 238 Plasticity 235 Polar satellite 196 Position vector and displacement 73 Potential energy of a spring 123 Potential energy 120 Power 128 Precession 143 Pressure gauge 253 Pressure of an ideal gas 328 Pressure 250 Principle of Conservation of Energy 128 Principle of moments 160 Progressive wave 373 Projectile motion 77 Projectile 77 Propagation constant 371 Pulse 369 Q Quasi-static process 310, 311 R Radiation 294 Radius of Gyration 164 Raman effect 11 Rarefactions 369 Ratio of specific heat capacities 334 Reaction time 51 Real gases 326 Rectilinear motion 39 Reductionism 2 Reflected wave 379 Reflection of waves 378 INDEX 311 Refracted wave 379 Refrigerator 313 Regelation 287 Relative velocity in two dimensions 76 Relative velocity 51 Resolution of vectors 69 Resonance 358 Restoring force 236, 350, 369 Reversible engine 316, 317 Reversible processes 315 Reynolds number 264 Rigid body 141 Rolling motion 173 Root mean square speed 329 Rotation 142 S S.H.M. (Simple Harmonic Motion) 343 Scalar-product 114 Scalars 65 Scientific Method 1 Second law of Thermodynamics 314 Shear modulus 242 Shearing strain 237 Shearing stress 237, 243 SI units 16 Significant figures 27 Simple pendulum 343, 353 Soap bubbles 268 Sonography 387 Sound 375 Specific heat capacity of Solids 308, 335 Specific heat capacity of Gases 333, 334 Specific heat capacity of Water 335 Specific heat capacity 285, 308 Speed of efflux 259 Speed of Sound 375, 376 Speed of Transverse wave 375, 376 on a stretched string Sphygmomanometer 277 Spring constant 352, 355 Standing waves 380 Stationary waves 382 Steady flow 257 Stethoscope 281 Stokes' law 263 Stopping distance 50 Strain 236 Streamline flow 257, 258 Streamline 257, 258 Stress 236 Stress-strain curve 238 Stretched string 374 Sublimation 294 Subtraction of vectors 67 Superposition principle 378 Surface energy 265 Surface tension 265 Symmetry 146 System of units 16 Systolic pressure 277 T Temperature 279 Tensile strength 238 Tensile stress 236 Terminal velocity 264 Theorem of parallel axes 167 Theorem of perpendicular axes 165 Thermal conductivity 291 Thermal equilibrium 304 Thermal expansion 281 Thermal stress 284 Thermodynamic processes 310 Thermodynamic state variables 309 Thermodynamics 3, 303 Time of flight 78 Torque 154 Torricelli's Law 259, 260 Trade wind 294 Transmitted wave 379 Travelling wave 380 Triangle law of addition of vectors 66 Triple point 288 Trough 371 Tune 384 Turbulent flow 258, 259 U Ultimate strength 238 Ultrasonic waves 387 Unification of Forces 10 Unified Atomic Mass Unit 21 Uniform circular motion 79 Uniform Motion 41 Uniformly accelerated motion 47 Unit vectors 70 V Vane 356 Vaporisation 288 Vector-product 151 Vectors 66 Velocity amplitude 349 Venturi meter 260 Vibration 341 Viscosity 262 Volume expansion 281 Volume Strain 238 W Wave equation 374 Wavelength 372 Wave speed 374 312 PHYSICS Waves 368 Waxing and waning of sound 385 Weak nuclear force 9 Weightlessness 197 Work done by variable force 118 Work 116 Work-Energy Theorem 116 Working substance 313 Y Yield Point 238 Yield strength 238 Young's modulus 239 Z Zeroth law of Thermodynamics 305 313 INDEX NOTES. Beats tell you the size of the frequency gap, not which string is higher — you find that out by seeing which way the gap moves when you change one string. Step 1: Use the beat frequency to get the two possible values of B's frequency. When two sources of close frequencies f A and f B sound together, the beat frequency is f beat = f A - f B Here f A = 324 Hz and f beat = 6 Hz, so 324 - f B = 6 This gives two candidates: f B = 324 + 6 = 330 Hz or f B = 324 - 6 = 318 Hz Beats alone cannot tell you which one is correct — a beat frequency only measures the magnitude of the gap, never its sign. That is exactly why the problem gives you a second clue: what happens when the tension in A is changed. Step 2: Relate tension to frequency for a stretched string. For a string under tension T, the fundamental frequency is f ∝ √ T So reducing the tension in A lowers its frequency — f A moves down from 324 Hz to some new value f A' < 324 Hz. This is the step people skip: they remember "tension changed" but forget it tells you the direction the frequency moved, and that direction is what breaks the tie between the two candidate values of B. Step 3: Test each candidate against the new (smaller) beat frequency. If f B = 330 Hz (B above A): lowering f A below 324 Hz pulls A further away from 330 Hz, so the gap f A' - 330 would grow larger than 6 Hz — the beat frequency would increase. This contradicts the given fact that the beats dropped to 3 Hz, so this value of f B is ruled out. If f B = 318 Hz (B below A): lowering f A from 324 Hz moves A closer to 318 Hz, so the gap f A' - 318 shrinks below 6 Hz — the beat frequency decreases, exactly as observed (it falls to 3 Hz). This is consistent. Step 4: Conclusion. Only f B = 318 Hz is consistent with the beat frequency falling from 6 Hz to 3 Hz when A's tension — and hence A's frequency — is reduced. Answer: The frequency of string B is 318 Hz.

Question 19

Explain why (or how): in a sound wave, a displacement node is a pressure antinode and vice versa, bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”, a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (e) the shape of a pulse gets distorted during propagation in a dispersive medium.

Accepted Answer

(a) A pressure antinode sits exactly where the displacement is zero, because pressure change depends on how fast displacement varies with position, not on the displacement itself. In a sound wave the excess pressure is related to the displacement s(x,t) through Hooke's law for a volume element, using the bulk modulus B: p(x,t) = -B (∂ s)/(∂ x) For a standing sound wave (as in an air column) the displacement pattern is s(x,t) = s 0 sin (kx) cos (ω t) Differentiating with respect to x gives the pressure variation p(x,t) = -B s 0 k cos (kx) cos (ω t) A displacement node is where sin (kx) = 0, so s = 0 there for all time — but at exactly that x, cos (kx) = ± 1, which is the largest possible value. So the pressure term is at its maximum: the displacement node is a pressure antinode. A displacement antinode is where sin (kx) = ± 1 (maximum swing of the air layer), but there cos (kx) = 0, so p = 0 always: the displacement antinode is a pressure node. The physical picture: at a displacement node, neighbouring layers on either side move toward or away from that point together, so the air gets alternately squeezed and stretched hardest right there — maximum compression/rarefaction, i.e. maximum pressure swing. At a displacement antinode, the whole local region moves back and forth in step, so no layer is compressed relative to its neighbour — pressure stays at atmospheric value. (b) Bats "see" with sound because a short, high-frequency pulse behaves like a ray, bounces back as an echo, and timing plus stereo comparison of that echo reconstructs the whole scene. Bats emit pulses of ultrasonic sound, typically with frequency in the range of tens of kHz to over 100 kHz — well above the human hearing limit of about 20 kHz. Using v = fλ with the speed of sound in air v ≈ 340 m/s, such high frequencies correspond to wavelengths of only a few millimetres, comparable to or smaller than the insects and obstacles the bat needs to detect. Because the wavelength is so short, the sound reflects off objects almost like light reflects off a mirror (diffraction, which would otherwise bend the wave around small obstacles, is negligible when λ size of the obstacle). This lets the returning echo carry sharp, direction-specific information rather than being smeared out. - Distance: the bat measures the time delay Δ t between emitting the pulse and receiving its echo. Since the sound travels to the obstacle and back, the distance is d = (v Δ t)/2 - Direction: the bat compares the echo's arrival time and intensity at its two ears; the tiny difference between the two ears (interaural time/intensity difference) tells it which way the obstacle lies. - Nature and size: the strength, frequency shift (from the Doppler effect if the target is moving), and detailed structure of the returning echo depend on the obstacle's size, shape and surface texture, letting the bat's brain build up a picture of what it is. This active use of reflected high-frequency sound to locate objects is called echolocation — the same principle used by SONAR and RADAR. (c) Pitch is set by frequency, but the "quality" of a note is set by which overtones ride along with it — and that is what lets the ear tell a violin from a sitar even at the same pitch. A stretched string does not vibrate purely at its fundamental frequency; it vibrates simultaneously at the fundamental and a set of overtones (harmonics), and the actual sound wave is the superposition of all of these: s(x,t) = Σ n A n sin (k n x) cos (ω n t + φ n) If two instruments are tuned so their fundamental frequency ω 1 is the same, the two notes have the same pitch. But the relative amplitudes A n and phases φ n of the higher harmonics depend on how the string is set into motion (bowed continuously, as in a violin, versus plucked, as in a sitar) and on the resonant body of the instrument that shapes the sound. These differences change the overall shape (waveform) of the resulting pressure wave even though its repetition rate (frequency) is identical. The human ear and brain are sensitive not just to the fundamental frequency but to this entire waveform shape, a property called timbre or quality. That is why a violin note and a sitar note of the same frequency still sound distinguishably different. (d) A transverse wave needs the medium to resist being sheared sideways, and gases simply do not resist that — they only resist being compressed. A wave can only exist where a restoring force brings a displaced element of the medium back toward equilibrium. - A transverse wave requires layers of the medium to resist sliding relative to one another (a shearing deformation). This restoring force exists only if the medium has a nonzero shear modulus (modulus of rigidity), η, which measures resistance to a change of shape at constant volume. - A longitudinal wave requires the medium to resist a change in volume (compression/rarefaction). This restoring force comes from the bulk modulus, B. Solids possess both a shear modulus and a bulk modulus — they hold their shape as well as their volume — so solids can carry both transverse waves (like the S-waves of an earthquake) and longitudinal waves (like the P-waves). Gases (and liquids) have no rigidity: they flow and take the shape of their container, offering zero resistance to a shearing stress, so their shear modulus is effectively zero. They do resist being compressed, so B ≠ 0. With no restoring force available for a sideways shear, a gas cannot sustain a transverse wave — it can only carry the longitudinal compression waves we hear as sound. (e) A pulse is a mixture of many wavelengths, and it only keeps its shape if all of them travel at the same speed — which a dispersive medium does not allow. By Fourier's theorem, a localized pulse — unlike a single-frequency sine wave — is actually a superposition of sinusoidal waves spanning a whole range of wavelengths (and hence frequencies): s(x,t) = ∫ A(k) cos (kx-ω(k) t) dk In a non-dispersive medium, every one of these component waves travels at the same phase speed v = ω/k = constant, independent of k. Their relative phase relationship never changes as they propagate, so the sum of all the components — the pulse — simply translates along at speed v without changing shape. In a dispersive medium, the relationship between ω and k is not a straight line through the origin, so v = ω/k depends on wavelength: different frequency components travel at different phase speeds. As the pulse moves forward, the faster components pull ahead of the slower ones, so the phase relationships among them keep changing with time. The sum of the components — the actual shape of the pulse — therefore spreads out and changes as it travels; it does not reproduce its original shape at a later position, and its envelope moves at the group velocity v g = dω/dk, which is generally different from the individual phase velocities. Answer: (a) A displacement node is where the strain (and hence pressure variation) is maximum, so it is a pressure antinode, and vice versa. (b) Bats emit ultrasonic pulses and use the echo's delay time (for distance) and the difference between the two ears (for direction), with the echo's strength and frequency shift revealing size and nature — this is echolocation. (c) Frequency fixes the pitch, but the amplitudes and phases of the accompanying overtones (the waveform, or timbre) differ between a bowed violin string and a plucked sitar string, letting the ear tell them apart. (d) Only solids have a nonzero shear modulus, so only solids can support the sideways restoring force a transverse wave needs; gases have only a bulk modulus, so they support just longitudinal waves. (e) A pulse is a superposition of many wavelengths; in a dispersive medium these travel at different phase speeds, so their relative phases drift and the pulse shape distorts as it propagates.

NCERT Solutions · Class 11 Physics Waves

Exercises 14.1–14.10 (part 1 of 2)

Exercise 14.1

Exercise 14.2

Exercise 14.3

Exercise 14.4

Exercise 14.5

Exercise 14.6

Exercise 14.7

Exercise 14.8

Exercise 14.9

Exercise 14.10

Exercises 14.11–14.19 (part 2 of 2)

Exercise 14.11

Exercise 14.12

Exercise 14.13

Exercise 14.14

Exercise 14.15

Exercise 14.16

Exercise 14.17

Exercise 14.18

Exercise 14.19