NCERT Solutions · Class 11 Physics Waves

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 $\displaystyle 1$: Find the linear mass density.Linear mass density \(\displaystyle \mu \) is the mass per unit length of the string: \[\mu = \frac{m}{L} \] where \(\displaystyle m = 2.50 \) kg is the mass of the string and \(\displaystyle L = 20.0 \) m is its length.\[\mu = \frac{2.50\ \text{kg}}{20.0\ \text{m}} = 0.125\ \text{kg/m} \]Step $\displaystyle 2$: Find the wave speed.The speed of a transverse wave on a stretched string is \[v = \sqrt{\frac{T}{\mu}} \] where \(\displaystyle T = 200 \) N is the tension in the string. This formula comes from applying Newton's second law to a small element of the string; \(\displaystyle T \) supplies the restoring force and \(\displaystyle \mu \) supplies the inertia — a taut, light string carries the jerk fast, a slack, heavy one carries it slowly.\[v = \sqrt{\frac{200\ \text{N}}{0.125\ \text{kg/m}}} = \sqrt{1600\ \text{m}^2/\text{s}^2} = 40.0\ \text{m/s} \]A step people trip on here: this \(\displaystyle v \) 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 $\displaystyle 3$: Find the time to cross the string.The disturbance travels the full length \(\displaystyle L \) of the string at this speed, so the time taken is \[t = \frac{L}{v} = \frac{20.0\ \text{m}}{40.0\ \text{m/s}} = 0.500\ \text{s} \]All three given quantities ($\displaystyle 2.50$ kg, $\displaystyle 200$ N, $\displaystyle 20.0$ m) carry three significant figures, so the answer is kept to three significant figures: \(\displaystyle 0.500 \) s.Answer: The disturbance takes $\displaystyle 0.500$ s to reach the other end of the string.

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 $\displaystyle 1$ — Time for the stone to fall $\displaystyle 300$ m.The stone starts from rest and falls freely, so use the second equation of motion for uniformly accelerated motion: \[h = u t_1 + \frac{1}{2}g t_1^2 \] Here \(\displaystyle u = 0\) (dropped, not thrown), \(\displaystyle h = 300\ \text{m}\) is the height of the tower, \(\displaystyle g = 9.8\ \text{m s}^{-2}\) is the acceleration due to gravity, and \(\displaystyle t_1\) is the time to reach the water.Since \(\displaystyle u=0\): \[h = \frac{1}{2}g t_1^2 \quad\Rightarrow\quad t_1 = \sqrt{\frac{2h}{g}} \]Substituting values: \[t_1 = \sqrt{\frac{2 \times 300\ \text{m}}{9.8\ \text{m s}^{-2}}} = \sqrt{\frac{600}{9.8}\ \text{s}^2} = \sqrt{61.22\ \text{s}^2} \] \[t_1 = 7.825\ \text{s} \]Step $\displaystyle 2$ — 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 $\displaystyle 300$ m at a constant speed (sound does not accelerate the way a falling body does). Using \[\text{speed} = \frac{\text{distance}}{\text{time}} \quad\Rightarrow\quad t_2 = \frac{h}{v_{\text{sound}}} \] where \(\displaystyle v_{\text{sound}} = 340\ \text{m s}^{-1}\).\[t_2 = \frac{300\ \text{m}}{340\ \text{m s}^{-1}} = 0.8824\ \text{s} \]Step $\displaystyle 3$ — 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 \[T = t_1 + t_2 = 7.825\ \text{s} + 0.8824\ \text{s} = 8.707\ \text{s} \]Rounding for significant figures: the given data (\(\displaystyle g = 9.8\), \(\displaystyle v = 340\)) carry two significant figures, so the final answer should also be quoted to two significant figures. Rounding \(\displaystyle 8.707\ \text{s}\) to two significant figures gives \(\displaystyle 8.7\ \text{s}\). Do not round \(\displaystyle t_1\) or \(\displaystyle t_2\) separately before adding — that is a common mistake that can shift the last digit.Answer: The splash is heard at the top $\displaystyle 8.7$ s after the stone is dropped.

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\[v = \sqrt{\dfrac{T}{\mu}} \]where \(\displaystyle T\) is the tension in the wire and \(\displaystyle \mu\) is its linear mass density (mass per unit length).Step $\displaystyle 1$: Find the linear mass density.\[\mu = \dfrac{m}{L} = \dfrac{2.10\ \text{kg}}{12.0\ \text{m}} = 0.175\ \text{kg}\,\text{m}^{-1} \]Step $\displaystyle 2$: 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 \(\displaystyle 20^{\circ}\)C, \(\displaystyle v = 343\ \text{m}\,\text{s}^{-1}\). 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,\[v^{2} = \dfrac{T}{\mu} \quad\Rightarrow\quad T = \mu v^{2} \]Step $\displaystyle 3$: Substitute the numbers.\[T = (0.175\ \text{kg}\,\text{m}^{-1}) \times (343\ \text{m}\,\text{s}^{-1})^{2} \]\[T = 0.175 \times 117649\ \text{kg}\,\text{m}\,\text{s}^{-2} \]\[T = 20588.575\ \text{N} \]Step $\displaystyle 4$: Round to the correct number of significant figures.Every measured quantity here — \(\displaystyle 2.10\ \text{kg}\), \(\displaystyle 12.0\ \text{m}\), and \(\displaystyle 343\ \text{m}\,\text{s}^{-1}\) — 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.\[T \approx 2.06 \times 10^{4}\ \text{N} \]This is a large tension (about $\displaystyle 2100$ kgf, or the weight of roughly two tonnes) — a reasonable size for a taut steel wire, since the wave speed required ($\displaystyle 343$ m/s) is very high for a wire this light.Answer: \(\displaystyle T \approx 2.06 \times 10^{4}\ \text{N}\) (about $\displaystyle 20600$ N).

The formula \(\displaystyle v = \sqrt{\dfrac{\gamma P}{\rho}} \) hides temperature and composition inside \(\displaystyle P \) and \(\displaystyle \rho \) — pull them out using the ideal gas law, and all three parts fall out together.Here \(\displaystyle v \) is the speed of sound, \(\displaystyle \gamma = C_p/C_v \) is the ratio of specific heats of the gas, \(\displaystyle P \) is the pressure, and \(\displaystyle \rho \) is the density of the gas.For \(\displaystyle n \) moles of an ideal gas occupying volume \(\displaystyle V \), the gas law gives \[PV = nRT \] where \(\displaystyle R \) is the universal gas constant and \(\displaystyle T \) is the absolute temperature. If \(\displaystyle M \) is the molar mass of the gas and \(\displaystyle m = nM \) is the mass of gas in that volume, then the density is \(\displaystyle \rho = m/V = nM/V \), so \(\displaystyle n/V = \rho/M \). Substituting into the gas law, \[P = \frac{n}{V}RT = \frac{\rho RT}{M} \quad\Rightarrow\quad \frac{P}{\rho} = \frac{RT}{M}. \] Putting this into the speed formula, \[v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T}{M}}. \] This rewritten form is the one to use for all three parts, since it separates out \(\displaystyle T \) and \(\displaystyle M \) explicitly.(a) Independent of pressure. At a fixed temperature, the ratio \(\displaystyle P/\rho \) equals \(\displaystyle RT/M \), which does not contain \(\displaystyle P \) at all — it depends only on \(\displaystyle T \) and on which gas it is. Physically: if you compress air at constant temperature, Boyle's law says \(\displaystyle P \propto \rho \) (density rises in the same proportion as pressure, since the same mass is squeezed into less volume), so the ratio \(\displaystyle P/\rho \) stays fixed even though \(\displaystyle P \) itself has changed. Since \(\displaystyle v = \sqrt{\gamma P/\rho} \) depends only on this ratio (and on \(\displaystyle \gamma \), which is also unchanged for a given gas at a given temperature), \(\displaystyle v \) does not change when \(\displaystyle P \) alone is changed at constant \(\displaystyle T \).(b) Increases with temperature. From \(\displaystyle v = \sqrt{\gamma R T/M} \), for a fixed gas (fixed \(\displaystyle \gamma \), \(\displaystyle R \), \(\displaystyle M \)), \[v \propto \sqrt{T}. \] Raising the temperature directly raises \(\displaystyle v \) — 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 \(\displaystyle v \) by raising \(\displaystyle T \) itself in the formula, while humidity (part c) raises \(\displaystyle v \) by lowering \(\displaystyle M \).(c) Increases with humidity. Moist air is a mixture of dry air (mean molar mass about \(\displaystyle 28.9\ \text{g/mol} \), mostly \(\displaystyle \text{N}_2 \) and \(\displaystyle \text{O}_2 \)) with water vapour (molar mass \(\displaystyle 18\ \text{g/mol} \)) 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 \(\displaystyle \text{N}_2 \)/\(\displaystyle \text{O}_2 \) molecules mole-for-mole, the average molar mass \(\displaystyle M \) of the mixture drops below that of dry air, even though \(\displaystyle T \) and \(\displaystyle P \) are unchanged. Since \[v = \sqrt{\frac{\gamma R T}{M}}, \] and \(\displaystyle M \) sits in the denominator, a smaller \(\displaystyle M \) gives a larger \(\displaystyle v \). 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 \(\displaystyle P/\rho = RT/M \) is independent of \(\displaystyle P \) (Boyle's law makes \(\displaystyle \rho \) rise in step with \(\displaystyle P \)), so \(\displaystyle v \) does not depend on pressure. (b) \(\displaystyle v = \sqrt{\gamma RT/M} \propto \sqrt{T} \), so \(\displaystyle v \) increases as temperature rises. (c) Water vapour lowers the mean molar mass \(\displaystyle M \) of humid air compared to dry air, and since \(\displaystyle v \propto 1/\sqrt{M} \), the speed of sound increases with humidity.

Being a function of \(\displaystyle x \pm vt\) 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 \(\displaystyle x \pm vt\) matters at all. A disturbance that keeps its shape while sliding along the \(\displaystyle x\)-axis with speed \(\displaystyle v\) must depend on position and time only through \(\displaystyle x - vt\) (shape moving in the \(\displaystyle +x\) direction) or \(\displaystyle x + vt\) (moving in the \(\displaystyle -x\) direction). Here \(\displaystyle y\) is the displacement of the medium, \(\displaystyle x\) the position of a point of the medium, \(\displaystyle t\) the time, and \(\displaystyle v\) the wave speed.Now the converse. Put \(\displaystyle u = x \pm vt\) and let \(\displaystyle y = f(u)\) be any twice-differentiable function. By the chain rule,\[\frac{\partial y}{\partial x} = f'(u),\qquad \frac{\partial^2 y}{\partial x^2} = f''(u),\] \[\frac{\partial y}{\partial t} = \pm v\, f'(u),\qquad \frac{\partial^2 y}{\partial t^2} = v^2 f''(u),\]so every such \(\displaystyle f\) obeys the wave equation\[\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\,\frac{\partial^2 y}{\partial t^2}.\]That is exactly why the converse fails as a physical statement: the form \(\displaystyle f(x\pm vt)\) 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: \(\displaystyle y\) is a real displacement of real matter, so it must be A pulse should also die away, \(\displaystyle y \to 0\) as \(\displaystyle |x| \to \infty\), 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 — \(\displaystyle (x-vt)^2\) is in \(\displaystyle \mathrm{m^2}\), \(\displaystyle 1/(x+vt)\) in \(\displaystyle \mathrm{m^{-1}}\), and a logarithm needs a dimensionless argument, which is what \(\displaystyle x_0\) 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) \(\displaystyle y = (x - vt)^2\). It is a function of \(\displaystyle x - vt\), so it passes the form test. But it is a parabola whose minimum sits at \(\displaystyle x = vt\) and which grows without limit on either side: at \(\displaystyle t = 0\), \(\displaystyle y = x^2\), so at \(\displaystyle x = 10\ \mathrm{m}\), \(\displaystyle y = 100\); at \(\displaystyle x = 100\ \mathrm{m}\), \(\displaystyle y = 10^4\); and \(\displaystyle y \to \infty\) as \(\displaystyle |x - vt| \to \infty\). The displacement is unbounded, and the medium is nowhere left undisturbed except at the single point \(\displaystyle x = vt\) — the opposite of a pulse. Not an acceptable travelling wave.(b) \(\displaystyle y = \log\!\left[\dfrac{x + vt}{x_0}\right]\). Again a function of \(\displaystyle x + vt\), so the form test passes, and it fails everything else: Not an acceptable travelling wave.(c) \(\displaystyle y = \dfrac{1}{x + vt}\). This is a function of \(\displaystyle x + vt\), and it is the only one of the three that behaves well far away: \(\displaystyle y \to 0\) as \(\displaystyle |x + vt| \to \infty\), 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, \(\displaystyle x + vt = 0\), i.e.\[x = -vt.\]That is a genuine point of the medium at every instant — for \(\displaystyle v = 2\ \mathrm{m\,s^{-1}}\) and \(\displaystyle t = 3\ \mathrm{s}\) it is the point \(\displaystyle x = -6\ \mathrm{m}\) — and there \(\displaystyle y \to \pm\infty\). The singularity is not at some unreachable place; it travels along the string in the \(\displaystyle -x\) direction with speed \(\displaystyle v\), 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: \(\displaystyle 1/(x+vt)\) is not finite everywhere, because it diverges on the line \(\displaystyle x = -vt\). Substituting any \(\displaystyle t\) 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 \(\displaystyle x = -vt\) — the reasoning, not the key, is what the question is asking for.Answer: No, the converse is not true — depending on \(\displaystyle x\) and \(\displaystyle t\) only through \(\displaystyle x \pm vt\) is necessary but not sufficient; the function must in addition remain finite (and defined) for all \(\displaystyle x\) and \(\displaystyle t\). None of the three qualifies: (a) \(\displaystyle (x-vt)^2\) grows without bound as \(\displaystyle |x-vt| \to \infty\); (b) \(\displaystyle \log[(x+vt)/x_0]\) is undefined for \(\displaystyle x+vt<0\) and diverges at \(\displaystyle x+vt = 0\) and as \(\displaystyle x+vt \to \infty\); (c) \(\displaystyle 1/(x+vt)\), although it decays at large distance, becomes infinite at \(\displaystyle x = -vt\). NCERT's printed answer accepting (c) contradicts the finiteness condition it itself states.

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 $\displaystyle 1000$ kHz call. What changes from one medium to the other is the speed of sound, and since\[v = f\lambda \quad\Rightarrow\quad \lambda = \frac{v}{f} \]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 numbersConvert the frequency to hertz first: \[f = 1000\ \text{kHz} = 1000 \times 10^{3}\ \text{Hz} = 1 \times 10^{6}\ \text{Hz} \]Given: \[v_{\text{air}} = 340\ \text{m s}^{-1}, \qquad v_{\text{water}} = 1486\ \text{m s}^{-1} \](a) The reflected soundThe 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\[\lambda_{\text{reflected}} = \frac{v_{\text{air}}}{f} = \frac{340\ \text{m s}^{-1}}{1 \times 10^{6}\ \text{Hz}} \]\[\lambda_{\text{reflected}} = 3.4 \times 10^{-4}\ \text{m} \]Keeping two significant figures (matching the two figures in $\displaystyle 340$ m s\(\displaystyle ^{-1}\)), this stays exactly \(\displaystyle 3.4\times10^{-4}\) m — no further rounding needed.(b) The transmitted soundThe transmitted wave now moves through water at $\displaystyle 1486$ m s\(\displaystyle ^{-1}\), with the frequency unchanged at \(\displaystyle 1\times10^{6}\) Hz:\[\lambda_{\text{transmitted}} = \frac{v_{\text{water}}}{f} = \frac{1486\ \text{m s}^{-1}}{1 \times 10^{6}\ \text{Hz}} \]\[\lambda_{\text{transmitted}} = 1.486 \times 10^{-3}\ \text{m} \]Rounding once, at the end, to three significant figures:\[\lambda_{\text{transmitted}} \approx 1.49 \times 10^{-3}\ \text{m} \]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 \(\displaystyle = 3.4 \times 10^{-4}\ \text{m}\); transmitted wavelength \(\displaystyle \approx 1.49 \times 10^{-3}\ \text{m}\) (frequency stays at \(\displaystyle 1\times10^{6}\) Hz in both cases; only the medium's speed changes the wavelength).

Wavelength links wave speed and frequency through \(\displaystyle v = f\lambda \), so \(\displaystyle \lambda = v/f\).The wave equation \(\displaystyle v = f\lambda \) holds for any travelling wave, where \(\displaystyle v\) is the speed of the wave in the medium, \(\displaystyle f\) is its frequency, and \(\displaystyle \lambda\) is its wavelength. Rearranging for the wavelength gives\[\lambda = \frac{v}{f} \]Convert both quantities to SI units before substituting.The speed of sound in the tissue is \[v = 1.7\ \text{km s}^{-1} = 1.7 \times 10^{3}\ \text{m s}^{-1} \]The operating frequency of the scanner is \[f = 4.2\ \text{MHz} = 4.2 \times 10^{6}\ \text{Hz} \]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 \(\displaystyle \lambda = v/f\).\[\lambda = \frac{1.7 \times 10^{3}\ \text{m s}^{-1}}{4.2 \times 10^{6}\ \text{Hz}} \]\[\lambda = 0.404761\ldots \times 10^{-3}\ \text{m} = 4.04761\ldots \times 10^{-4}\ \text{m} \]Round to significant figures.Both given quantities, $\displaystyle 1.7$ and $\displaystyle 4.2$, carry only two significant figures, so the answer must also be quoted to two significant figures:\[\lambda \approx 4.0 \times 10^{-4}\ \text{m} = 0.40\ \text{mm} \]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: \(\displaystyle \lambda \approx 4.0 \times 10^{-4}\ \text{m} = 0.40\ \text{mm}\)

A travelling wave has \(\displaystyle x\) and \(\displaystyle t\) tangled together inside one sine argument; a stationary wave keeps them apart, as a product like \(\displaystyle 2A\sin(kx)\cos(\omega t)\). Here\[y(x,t) = 3.0\sin(36t + 0.018x + \pi/4) \]has \(\displaystyle x\) and \(\displaystyle t\) added together inside a single sine, so this is a travelling wave, not a stationary one.(a) Speed and directionMatch the given equation to the general travelling-wave form \(\displaystyle y = A\sin(kx + \omega t + \phi)\), reading off \[A = 3.0\ \text{cm}, \qquad k = 0.018\ \text{rad/cm}, \qquad \omega = 36\ \text{rad/s}, \qquad \phi = \pi/4. \]The sign in front of \(\displaystyle t\) is the part people trip over: a wave moving in the \(\displaystyle +x\) direction is written \(\displaystyle \sin(kx - \omega t)\), while a wave moving in the \(\displaystyle -x\) direction is written \(\displaystyle \sin(kx + \omega t)\) — it is the relative sign between the \(\displaystyle kx\) and \(\displaystyle \omega t\) terms that fixes the direction, not the sign of \(\displaystyle k\) or \(\displaystyle \omega\) on their own. Since this equation has \(\displaystyle kx\) and \(\displaystyle \omega t\) with the same sign, the wave travels in the \(\displaystyle -x\) direction. Because positive \(\displaystyle x\) is defined as left to right, this wave moves from right to left.The speed of a travelling wave is \(\displaystyle v = \omega/k\), where \(\displaystyle \omega\) is the angular frequency and \(\displaystyle k\) is the angular wave number: \[v = \frac{\omega}{k} = \frac{36\ \text{rad/s}}{0.018\ \text{rad/cm}} = 2.0\times10^{3}\ \text{cm/s} = 20\ \text{m/s}. \](b) Amplitude and frequencyThe amplitude is simply the coefficient of the sine, already read off above: \(\displaystyle A = 3.0\ \text{cm}\).The frequency comes from \(\displaystyle \omega = 2\pi f\): \[f = \frac{\omega}{2\pi} = \frac{36\ \text{rad/s}}{2\pi} = 5.7296\ldots\ \text{Hz}. \] Rounding once, to two significant figures (matching the $\displaystyle 36$ and $\displaystyle 0.018$ in the data), \(\displaystyle f \approx 5.7\ \text{Hz}\).(c) Initial phase at the originThe phase of the wave is \(\displaystyle \theta(x,t) = 36t + 0.018x + \pi/4\). "At the origin" and "initial" mean \(\displaystyle x = 0\) and \(\displaystyle t = 0\): \[\theta(0,0) = 36(0) + 0.018(0) + \pi/4 = \pi/4. \](d) Least distance between two successive crestsSuccessive crests are one full spatial cycle apart, so this distance is the wavelength \(\displaystyle \lambda\), related to the wave number by \(\displaystyle k = 2\pi/\lambda\): \[\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.018\ \text{rad/cm}} = 349.07\ \text{cm}. \] Converting and rounding to two significant figures, \(\displaystyle \lambda \approx 3.5\ \text{m}\).Answer: It is a travelling wave, moving in the \(\displaystyle -x\) direction (right to left), with speed \(\displaystyle v = 20\ \text{m/s}\), amplitude \(\displaystyle A = 3.0\ \text{cm}\), frequency \(\displaystyle f \approx 5.7\ \text{Hz}\), initial phase at the origin \(\displaystyle \pi/4\), and least distance between successive crests (wavelength) \(\displaystyle \lambda \approx 3.5\ \text{m}\).

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 \[y(x,t) = 3.0\sin(36t + 0.018x + \pi/4)\ \text{cm}, \qquad x\ \text{in cm},\ t\ \text{in s}. \] Here the angular frequency is \(\displaystyle \omega = 36\ \text{rad s}^{-1}\), the angular wave number is \(\displaystyle k = 0.018\ \text{rad cm}^{-1}\), the amplitude is \(\displaystyle A = 3.0\ \text{cm}\), and \(\displaystyle \phi_0=\pi/4\) is the phase at the origin.Displacement at each of the three points. To get \(\displaystyle y\) versus \(\displaystyle t\) at a fixed \(\displaystyle x\), just substitute that value of \(\displaystyle x\) into the wave equation — \(\displaystyle x\) then only adds a constant to the phase, it does not touch the \(\displaystyle t\)-dependence or the coefficient out front:\[x=0:\quad y_1(t) = 3.0\sin(36t + \pi/4)\ \text{cm} \] \[x=2\ \text{cm}:\quad y_2(t) = 3.0\sin\big(36t + \pi/4 + (0.018)(2)\big) = 3.0\sin(36t + 0.821)\ \text{cm} \] \[x=4\ \text{cm}:\quad y_3(t) = 3.0\sin\big(36t + \pi/4 + (0.018)(4)\big) = 3.0\sin(36t + 0.857)\ \text{cm} \] (using \(\displaystyle \pi/4 = 0.785\ \text{rad}\), so the extra phases are \(\displaystyle 0.018\times2=0.036\ \text{rad}\) and \(\displaystyle 0.018\times4=0.072\ \text{rad}\)).Shape of the graphs. Each of \(\displaystyle y_1(t)\), \(\displaystyle y_2(t)\), \(\displaystyle y_3(t)\) is a pure sine curve — a smooth, repeating up-and-down curve, not a straight line or a triangular wave — with: \[T = \frac{2\pi}{\omega} = \frac{2\pi}{36\ \text{rad s}^{-1}} \approx 0.17\ \text{s}, \] 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 \(\displaystyle x=2\) cm is displaced in time relative to the one for \(\displaystyle x=0\), and the curve for \(\displaystyle x=4\) cm is displaced by twice as much. This is because this wave travels in the \(\displaystyle -x\) direction (as found in Exercise $\displaystyle 14.8$: the term \(\displaystyle 0.018x\) has the same sign as \(\displaystyle 36t\), which is the signature of a \(\displaystyle -x\)-moving wave). A point farther along \(\displaystyle +x\) therefore feels the disturbance earlier than a point closer to the origin. Converting the extra phase to a time lead, \(\displaystyle \Delta t = kx/\omega\): \[\Delta t_{x=2} = \frac{0.036}{36}\ \text{s} = 1.0\times10^{-3}\ \text{s}, \qquad \Delta t_{x=4} = \frac{0.072}{36}\ \text{s} = 2.0\times10^{-3}\ \text{s}. \] So the graph at \(\displaystyle x=2\ \text{cm}\) is the same sine curve as at \(\displaystyle x=0\), just slid \(\displaystyle 1.0\ \text{ms}\) earlier on the time axis, and the graph at \(\displaystyle x=4\ \text{cm}\) is slid \(\displaystyle 2.0\ \text{ms}\) earlier still. (These shifts are tiny compared with the \(\displaystyle 0.17\ \text{s}\) 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 \(\displaystyle y\)–\(\displaystyle t\) graphs at \(\displaystyle x=0\), \(\displaystyle 2\) cm and \(\displaystyle 4\) cm are all identical sine curves of amplitude \(\displaystyle 3.0\ \text{cm}\) and period \(\displaystyle T\approx0.17\ \text{s}\); the curves for \(\displaystyle x=2\) cm and \(\displaystyle x=4\) cm are the same curve as at \(\displaystyle x=0\) shifted earlier in time by \(\displaystyle 1.0\times10^{-3}\,\text{s}\) and \(\displaystyle 2.0\times10^{-3}\,\text{s}\) 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 \(\displaystyle kx\)) that makes the disturbance appear to travel along the string.

At one instant, the phase difference between two points on a travelling wave depends only on how far apart they are: \(\displaystyle \Delta\phi = k\,\Delta x\), where \(\displaystyle k\) is the angular wave number.Start by putting the wave in the standard form \(\displaystyle y = a\cos(\omega t - kx + \phi_0)\), in which \(\displaystyle a\) is the amplitude, \(\displaystyle \omega\) the angular frequency, \(\displaystyle k\) the angular wave number and \(\displaystyle \phi_0\) the initial phase constant. Multiplying the \(\displaystyle 2\pi\) through the bracket,\[y(x,t) = 2.0\cos\!\big[\,2\pi(10t - 0.0080x + 0.35)\,\big] = 2.0\cos\!\big(20\pi\,t - 0.016\pi\,x + 0.70\pi\big)\ \text{cm} \]Reading off the coefficients:\[a = 2.0\ \text{cm},\qquad \omega = 20\pi\ \text{rad s}^{-1},\qquad k = 0.016\pi\ \text{rad cm}^{-1},\qquad \phi_0 = 0.70\pi\ \text{rad} \]The wavelength follows from \(\displaystyle k = 2\pi/\lambda\):\[\lambda = \frac{2\pi}{k} = \frac{2\pi\ \text{rad}}{0.016\pi\ \text{rad cm}^{-1}} = 125\ \text{cm} = 1.25\ \text{m} \]Watch the units here. The problem states that \(\displaystyle x\) is in centimetres, so \(\displaystyle k\) comes out in radians per centimetre. Every separation must be converted to centimetres before it is multiplied by \(\displaystyle k\); using metres makes the answer $\displaystyle 100$ 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 \(\displaystyle x\) is \(\displaystyle \phi(x) = \omega t - kx + \phi_0\). For two points \(\displaystyle x_1\) and \(\displaystyle x_2 = x_1 + \Delta x\) at the same instant \(\displaystyle t\),\[\Delta\phi = |\phi(x_2) - \phi(x_1)| = k\,\Delta x = \frac{2\pi}{\lambda}\,\Delta x \]The \(\displaystyle \omega t\) term and the constant \(\displaystyle \phi_0\) are the same for both points, so they cancel — which is why the answer does not depend on when you look.(a) Separation $\displaystyle 4$ m. Convert: \(\displaystyle \Delta x = 4\ \text{m} = 400\ \text{cm}\).\[\Delta\phi = (0.016\pi\ \text{rad cm}^{-1})(400\ \text{cm}) = 6.4\pi\ \text{rad} \approx 20\ \text{rad} \](b) Separation $\displaystyle 0.5$ m. Convert: \(\displaystyle \Delta x = 0.5\ \text{m} = 50\ \text{cm}\).\[\Delta\phi = (0.016\pi\ \text{rad cm}^{-1})(50\ \text{cm}) = 0.8\pi\ \text{rad} \approx 2.5\ \text{rad} \](c) Separation \(\displaystyle \lambda/2\). Here the separation is given as a fraction of the wavelength, so \(\displaystyle \lambda\) cancels and no number needs substituting:\[\Delta\phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{2} = \pi\ \text{rad} \approx 3.14\ \text{rad} \]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 \(\displaystyle 3\lambda/4\).\[\Delta\phi = \frac{2\pi}{\lambda}\cdot\frac{3\lambda}{4} = \frac{3\pi}{2}\ \text{rad} \approx 4.71\ \text{rad} \]Because \(\displaystyle \cos\) repeats every \(\displaystyle 2\pi\), a lag of \(\displaystyle 3\pi/2\) is the same state of motion as a lead of \(\displaystyle \pi/2\):\[\frac{3\pi}{2} - 2\pi = -\frac{\pi}{2} \]So the two points can equally be described as \(\displaystyle \pi/2\) rad out of step, with the far point leading rather than lagging. NCERT prints \(\displaystyle \pi/2\ \text{rad}\) 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 \(\displaystyle 6.4\pi\) would have collapsed to \(\displaystyle 0.4\pi\) by the same rule. Taken straight from \(\displaystyle \Delta\phi = k\,\Delta x\), and consistently with the first three parts, the value is \(\displaystyle 3\pi/2\) rad; quote \(\displaystyle \pi/2\) 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 \(\displaystyle 0.0080\) carries two significant figures, so the decimal values in (a) and (b) round to \(\displaystyle 20\) rad and \(\displaystyle 2.5\) rad. Parts (c) and (d) are exact multiples of \(\displaystyle \pi\) — the separations were given as fractions of \(\displaystyle \lambda\), so \(\displaystyle k\) cancels and its precision never enters.Answer: \(\displaystyle \Delta\phi = k\,\Delta x\) with \(\displaystyle k = 0.016\pi\ \text{rad cm}^{-1}\) and \(\displaystyle \lambda = 125\ \text{cm} = 1.25\ \text{m}\); (a) \(\displaystyle 6.4\pi\ \text{rad} \approx 20\ \text{rad}\); (b) \(\displaystyle 0.8\pi\ \text{rad} \approx 2.5\ \text{rad}\); (c) \(\displaystyle \pi\ \text{rad} \approx 3.14\ \text{rad}\); (d) \(\displaystyle 3\pi/2\ \text{rad} \approx 4.71\ \text{rad}\), equivalently \(\displaystyle \pi/2\) rad if the angle is reduced modulo \(\displaystyle 2\pi\), which is the form NCERT prints.