NCERT Solutions · Class 11 Physics Waves

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 \(\displaystyle x - vt\) (or \(\displaystyle x+vt\)).The given displacement is \[y(x,t) = 0.06\,\sin\!\left(\frac{2\pi x}{3}\right)\cos(120\pi t) \]Here the x-dependence and the t-dependence sit in two separate factors — \(\displaystyle \sin(2\pi x/3)\) fixes a shape along the string, and \(\displaystyle \cos(120\pi t)\) makes the amplitude of that fixed shape rise and fall in time. A travelling wave, by contrast, must be expressible as a function of \(\displaystyle x\) and \(\displaystyle t\) only through the combination \(\displaystyle x-vt\), so that the whole waveform slides along without changing shape. Because \(\displaystyle y(x,t)\) 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 \(\displaystyle \sin(2\pi x/3)=0\) 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 \(\displaystyle \sin A\cos B = \tfrac12\big[\sin(A+B)+\sin(A-B)\big]\) with \(\displaystyle A = \dfrac{2\pi x}{3}\), \(\displaystyle B = 120\pi t\):\[y(x,t) = 0.06\sin\!\left(\frac{2\pi x}{3}\right)\cos(120\pi t) = 0.03\sin\!\left(\frac{2\pi x}{3} - 120\pi t\right) + 0.03\sin\!\left(\frac{2\pi x}{3} + 120\pi t\right) \]A term of the form \(\displaystyle \sin(kx-\omega t)\) travels in the \(\displaystyle +x\) direction (to keep the argument fixed as \(\displaystyle t\) grows, \(\displaystyle x\) must grow too); a term of the form \(\displaystyle \sin(kx+\omega t)\) travels in the \(\displaystyle -x\) direction. So the standing wave is the sum of\[y_1 = 0.03\sin\!\left(\frac{2\pi x}{3} - 120\pi t\right)\text{ m, travelling in }+x,\qquad y_2 = 0.03\sin\!\left(\frac{2\pi x}{3} + 120\pi t\right)\text{ m, travelling in }-x, \]each of amplitude $\displaystyle 0.03$ m (half the standing-wave amplitude — this halving is exactly what superposition of two $\displaystyle 0.03$ m waves reconstructs).Reading off the wave number and angular frequency common to both: \(\displaystyle k = \dfrac{2\pi}{3}\ \text{rad/m}\), \(\displaystyle \omega = 120\pi\ \text{rad/s}\). These come straight from the given equation, so they carry no rounding error.Wavelength: \(\displaystyle k = \dfrac{2\pi}{\lambda} \Rightarrow \lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{2\pi/3} = 3\ \text{m}\)Frequency: \(\displaystyle \omega = 2\pi f \Rightarrow f = \dfrac{\omega}{2\pi} = \dfrac{120\pi}{2\pi} = 60\ \text{Hz}\)Speed: \(\displaystyle v = f\lambda = 60\ \text{Hz} \times 3\ \text{m} = 180\ \text{m/s}\)Both component waves — same \(\displaystyle \lambda\), same \(\displaystyle f\), same speed, opposite direction — have \(\displaystyle \lambda = 3\ \text{m}\), \(\displaystyle f = 60\ \text{Hz}\), \(\displaystyle v = 180\ \text{m/s}\).(c) The wave speed on a stretched string ties directly to its tension through \(\displaystyle v=\sqrt{T/\mu}\), where \(\displaystyle \mu\) is mass per unit length — not the total mass.\[\mu = \frac{\text{mass}}{\text{length}} = \frac{3.0\times10^{-2}\ \text{kg}}{1.5\ \text{m}} = 2.0\times10^{-2}\ \text{kg/m} \](A common slip here is plugging the string's total mass into \(\displaystyle v=\sqrt{T/m}\) directly — the formula needs mass per metre, so the division by length is not optional.)From \(\displaystyle v = \sqrt{T/\mu}\), squaring gives \(\displaystyle T = \mu v^2\):\[T = (2.0\times10^{-2}\ \text{kg/m})(180\ \text{m/s})^2 = (2.0\times10^{-2}\ \text{kg/m})(3.24\times10^{4}\ \text{m}^2/\text{s}^2) = 648\ \text{kg·m/s}^2 = 648\ \text{N} \]The mass ($\displaystyle 3.0$×$\displaystyle 10$⁻² kg) and length ($\displaystyle 1.5$ m) each carry only $\displaystyle 2$ significant figures, so \(\displaystyle \mu\) — and hence T — is good to $\displaystyle 2$ significant figures: rounding once, at the end, \(\displaystyle T \approx 6.5\times10^{2}\ \text{N}\).**Answer: (a) A stationary (standing) wave, since \(\displaystyle x\) and \(\displaystyle t\) appear in separate factors, not as \(\displaystyle x-vt\). (b) \(\displaystyle y = 0.03\sin(2\pi x/3 - 120\pi t) + 0.03\sin(2\pi x/3 + 120\pi t)\) m, i.e. two waves of amplitude $\displaystyle 0.03$ m travelling in opposite directions (+x and −x), each with \(\displaystyle \lambda = 3\ \text{m}\), \(\displaystyle f = 60\ \text{Hz}\), \(\displaystyle v = 180\ \text{m/s}\). (c) \(\displaystyle T = 648\ \text{N} \approx 6.5\times10^{2}\ \text{N}\) ($\displaystyle 2$ significant figures).

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\[y(x,t) = 0.06\,\sin\!\left(\frac{2\pi x}{3}\right)\cos(120\pi t)\ \text{m}, \qquad 0 \le x \le 1.5\ \text{m}. \]This has the standard stationary-wave form \(\displaystyle y = A(x)\cos(\omega t)\), where Every point on the string obeys \(\displaystyle y = A(x)\cos(\omega t)\) — the same function of time, only scaled by \(\displaystyle A(x)\). That single observation answers all three parts of (i).(i)(a) Frequency — yes, the same for all points.The time dependence is \(\displaystyle \cos(\omega t)\) with \(\displaystyle \omega = 120\pi\ \mathrm{rad\,s^{-1}}\), and \(\displaystyle \omega = 2\pi\nu\), so\[\nu = \frac{\omega}{2\pi} = \frac{120\pi\ \mathrm{rad\,s^{-1}}}{2\pi} = 60\ \mathrm{Hz}. \]Nothing in \(\displaystyle \cos(120\pi t)\) depends on \(\displaystyle x\), so every oscillating point of the string completes its cycle in the same time: all points vibrate at \(\displaystyle 60\ \mathrm{Hz}\). (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 \(\displaystyle A(x)\): if \(\displaystyle A(x) > 0\) the point moves with \(\displaystyle \cos(\omega t)\); if \(\displaystyle A(x) < 0\) it moves with \(\displaystyle -\cos(\omega t)\), i.e. exactly \(\displaystyle \pi\) out of phase. So the question becomes: does \(\displaystyle \sin(2\pi x/3)\) change sign anywhere on this string?Nodes are the points where \(\displaystyle A(x) = 0\):\[\sin\!\left(\frac{2\pi x}{3}\right) = 0 \ \Longrightarrow\ \frac{2\pi x}{3} = n\pi \ \Longrightarrow\ x = 1.5\,n\ \text{m}, \quad n = 0, 1, 2, \dots \]Inside the string \(\displaystyle 0 \le x \le 1.5\ \mathrm{m}\), that gives nodes at \(\displaystyle x = 0\) and \(\displaystyle x = 1.5\ \mathrm{m}\) — the two clamped ends — and nowhere else. Equivalently, \(\displaystyle \lambda = 2\pi/k = 3\ \mathrm{m}\), and \(\displaystyle L = 1.5\ \mathrm{m} = \lambda/2\): the string is vibrating in its fundamental mode, one single loop.With no interior node, \(\displaystyle \sin(2\pi x/3) > 0\) for every \(\displaystyle 0 < x < 1.5\ \mathrm{m}\). Every point therefore carries the same \(\displaystyle +\cos(120\pi t)\) 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 \(\displaystyle \pi\) 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 \(\displaystyle A(x) = 0.06\,\sin(2\pi x/3)\) m, which plainly depends on \(\displaystyle x\): it is \(\displaystyle 0\) at the nodes \(\displaystyle x = 0\) and \(\displaystyle x = 1.5\ \mathrm{m}\), and rises to its largest value \(\displaystyle 0.06\ \mathrm{m}\) at the antinode in the middle, \(\displaystyle x = 0.75\ \mathrm{m}\), where \(\displaystyle \sin(2\pi x/3) = \sin(\pi/2) = 1\). So the amplitude is not the same for all points.(ii) Amplitude of the point \(\displaystyle 0.375\ \mathrm{m}\) from one end.Measure \(\displaystyle x\) from the clamped end at \(\displaystyle x = 0\), so \(\displaystyle x = 0.375\ \mathrm{m}\), and substitute into \(\displaystyle A(x)\):\[A(0.375) = 0.06 \times \sin\!\left(\frac{2\pi \times 0.375}{3}\right)\ \mathrm{m} = 0.06 \times \sin\!\left(\frac{0.75\pi}{3}\right)\ \mathrm{m} = 0.06 \times \sin\!\left(\frac{\pi}{4}\right)\ \mathrm{m}. \]The angle here is in radians, not degrees — \(\displaystyle kx\) is always a radian measure, and \(\displaystyle \pi/4\ \mathrm{rad} = 45^\circ\). Using \(\displaystyle \sin(\pi/4) = 1/\sqrt{2}\):\[A(0.375) = \frac{0.06}{\sqrt{2}}\ \mathrm{m} = 0.042426\ldots\ \mathrm{m}. \]Rounding once, at the end, to the two significant figures carried by the data:\[A(0.375\ \mathrm{m}) = 0.042\ \mathrm{m} = 4.2\ \mathrm{cm}. \]A sanity check on the size: \(\displaystyle x = 0.375\ \mathrm{m}\) is halfway between the node at \(\displaystyle x = 0\) and the antinode at \(\displaystyle x = 0.75\ \mathrm{m}\), so the amplitude should be a good fraction of the antinode's \(\displaystyle 0.06\ \mathrm{m}\) but clearly below it — and \(\displaystyle 0.042\ \mathrm{m}\) is \(\displaystyle 1/\sqrt{2}\) of it, which is exactly what \(\displaystyle \sin 45^\circ\) demands.Answer: (i)(a) Yes — every oscillating point has the same frequency, \(\displaystyle \nu = 60\ \mathrm{Hz}\). (b) Yes — the string is vibrating in one loop (\(\displaystyle \lambda = 3\ \mathrm{m}\), \(\displaystyle L = \lambda/2\)), with nodes only at the two clamped ends, so \(\displaystyle 0.06\sin(2\pi x/3)\) never changes sign and all points except the nodes oscillate in phase. (c) No — the amplitude varies with position as \(\displaystyle A(x) = 0.06\sin(2\pi x/3)\) m, from \(\displaystyle 0\) at the nodes to \(\displaystyle 0.06\ \mathrm{m}\) at the antinode \(\displaystyle x = 0.75\ \mathrm{m}\). (ii) \(\displaystyle A = 0.06/\sqrt{2} = 0.042\ \mathrm{m}\) ($\displaystyle 4.2$ cm), to two significant figures.

Two tests decide every part — but only after a third one is passed. A function represents a travelling wave if \(\displaystyle x\) and \(\displaystyle t\) enter only through the single combination \(\displaystyle x-vt\) or \(\displaystyle x+vt\); it represents a stationary wave if it separates as (a shape in \(\displaystyle x\)) \(\displaystyle \times\) (an oscillation in \(\displaystyle t\)), 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 \(\displaystyle x\) and every \(\displaystyle t\), and it must satisfy the wave equation\[\frac{\partial^{2}y}{\partial x^{2}}=\frac{1}{v^{2}}\,\frac{\partial^{2}y}{\partial t^{2}}\]where \(\displaystyle v\) is the wave speed. A function that fails that is the "none at all" case.(a) \(\displaystyle y = 2\cos(3x)\,\sin(10t)\)This is already a product of a function of \(\displaystyle x\) alone and a function of \(\displaystyle t\) alone, which is the signature of a stationary wave. Comparing with \(\displaystyle y = 2a\cos kx\,\sin\omega t\):\[k = 3,\qquad \omega = 10 .\]Every particle performs simple harmonic motion of the same angular frequency \(\displaystyle \omega = 10\), with an amplitude \(\displaystyle 2\cos 3x\) that depends on where the particle sits. The nodes are the points that never move, \(\displaystyle \cos 3x = 0\):\[3x = (2n+1)\frac{\pi}{2}\quad\Rightarrow\quad x = (2n+1)\frac{\pi}{6},\qquad n = 0,1,2,\dots\]Using \(\displaystyle 2\sin A\cos B = \sin(A+B)+\sin(A-B)\) with \(\displaystyle A = 10t\), \(\displaystyle B = 3x\),\[y = \sin(10t+3x) + \sin(10t-3x),\]so it is the sum of two identical harmonic waves running in opposite directions with speed \(\displaystyle v = \omega/k = 10/3 \approx 3.3\) units. That is exactly how a stationary wave is made.Stationary wave.(b) \(\displaystyle y = 2\sqrt{x-vt}\)This is the part that traps people: \(\displaystyle x\) and \(\displaystyle t\) do appear only through \(\displaystyle x-vt\), 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. So there is no region of \(\displaystyle x\) and \(\displaystyle t\) in which this is a legitimate wave.None at all — an unacceptable function for any wave.(c) \(\displaystyle y = 3\sin(5x-0.5t) + 4\cos(5x-0.5t)\)Write \(\displaystyle \theta = 5x - 0.5t\) and combine the two terms using \(\displaystyle a\sin\theta + b\cos\theta = R\sin(\theta+\phi)\) with \(\displaystyle R = \sqrt{a^{2}+b^{2}}\) and \(\displaystyle \tan\phi = b/a\):\[R = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25} = 5,\qquad \phi = \tan^{-1}\!\left(\frac{4}{3}\right) = 0.927\ \text{rad}\ (53.1^{\circ}).\]\[y = 5\sin\!\left(5x - 0.5t + 0.927\right).\]It is a single sine of one argument, and that argument contains \(\displaystyle x\) and \(\displaystyle t\) only through\[5x - 0.5t = 5\left(x - 0.1\,t\right),\]which is the combination \(\displaystyle x - vt\). Comparing with \(\displaystyle y = A\sin(kx-\omega t+\phi)\):\[k = 5,\qquad \omega = 0.5,\qquad v = \frac{\omega}{k} = \frac{0.5}{5} = 0.1 .\]The minus sign between \(\displaystyle kx\) and \(\displaystyle \omega t\) means the pattern moves towards increasing \(\displaystyle x\) — a direction is part of this answer, not an optional extra. Amplitude \(\displaystyle 5\), speed \(\displaystyle 0.1\) in the units of the question (\(\displaystyle 0.1\ \mathrm{m\,s^{-1}}\) if \(\displaystyle x\) is in metres and \(\displaystyle t\) in seconds), quoted to the one significant figure the data carry.Travelling harmonic wave, moving in the \(\displaystyle +x\) direction.(d) \(\displaystyle y = \cos x\,\sin t + \cos 2x\,\sin 2t\)Take each term on its own. \(\displaystyle \cos x\,\sin t\) is separable, so it is a stationary wave with \(\displaystyle k_1 = 1,\ \omega_1 = 1\); \(\displaystyle \cos 2x\,\sin 2t\) is separable, so it is a stationary wave with \(\displaystyle k_2 = 2,\ \omega_2 = 2\). Both have the same speed \(\displaystyle \omega/k = 1\), 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:\[\frac{\partial^{2}y}{\partial x^{2}} = -\cos x\,\sin t - 4\cos 2x\,\sin 2t,\qquad \frac{\partial^{2}y}{\partial t^{2}} = -\cos x\,\sin t - 4\cos 2x\,\sin 2t,\]so \(\displaystyle \partial^{2}y/\partial x^{2} = \partial^{2}y/\partial t^{2}\), which is the wave equation with \(\displaystyle v = 1\). It is also bounded, \(\displaystyle |y| \le 2\). 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 \(\displaystyle t = 0\), \(\displaystyle y = \cos x\,\sin 0 + \cos 2x\,\sin 0 = 0\) for every \(\displaystyle x\). A travelling wave \(\displaystyle y = f(x-vt)\) just slides a fixed profile along, so if the profile is zero everywhere at one instant, \(\displaystyle f\) is identically zero and the wave is zero for all time. Here \(\displaystyle y\) is not zero at later times — at \(\displaystyle x = \pi/2\), for instance, \(\displaystyle y = -\sin 2t\). 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 \(\displaystyle 1\) and \(\displaystyle 2\). It is worth seeing why that phrasing is used rather than "a stationary wave" flat: the sum does not factorise as (function of \(\displaystyle x\))\(\displaystyle \times\)(function of \(\displaystyle t\)), and it has no permanent nodes — a node would need \(\displaystyle \cos x = 0\) and \(\displaystyle \cos 2x = 0\) at the same \(\displaystyle x\), but \(\displaystyle \cos 2x = 2\cos^{2}x - 1 = -1\) wherever \(\displaystyle \cos x = 0\). Each particle still oscillates about its mean position with a fixed period \(\displaystyle 2\pi\), 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 \(\displaystyle x=(2n+1)\pi/6\)); (b) none at all — \(\displaystyle 2\sqrt{x-vt}\) is imaginary for \(\displaystyle x<vt\) and unbounded for large \(\displaystyle x-vt\), so it is an unacceptable function for any wave; (c) travelling harmonic wave, \(\displaystyle y=5\sin(5x-0.5t+0.927)\), moving in the \(\displaystyle +x\) direction with speed \(\displaystyle v=\omega/k=0.1\) (\(\displaystyle 0.1\ \mathrm{m\,s^{-1}}\) for \(\displaystyle x\) in metres, \(\displaystyle t\) in seconds); (d) stationary — the superposition of two stationary waves of angular frequencies \(\displaystyle 1\) and \(\displaystyle 2\).

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 $\displaystyle 1$ — Find the length of the wire.Linear mass density is mass per unit length, \(\displaystyle \mu = M/L \), so the length is \[L = \frac{M}{\mu} \] With \(\displaystyle M = 3.5 \times 10^{-2}\ \text{kg} \) and \(\displaystyle \mu = 4.0 \times 10^{-2}\ \text{kg m}^{-1} \): \[L = \frac{3.5 \times 10^{-2}\ \text{kg}}{4.0 \times 10^{-2}\ \text{kg m}^{-1}} = 0.875\ \text{m} \]Step $\displaystyle 2$ — 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: \[L = \frac{\lambda}{2} \quad \Rightarrow \quad \lambda = 2L \] The wave speed is related to frequency and wavelength by \(\displaystyle v = f\lambda \), so \[v = f(2L) = 2Lf \] Substituting \(\displaystyle L = 0.875\ \text{m} \) and \(\displaystyle f = 45\ \text{Hz} \): \[v = 2 \times 0.875\ \text{m} \times 45\ \text{Hz} = 78.75\ \text{m s}^{-1} \] 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): \(\displaystyle v \approx 79\ \text{m s}^{-1} \).Step $\displaystyle 3$ — Tension in the string, from the wave-speed formula.The speed of a transverse wave on a stretched string is \(\displaystyle v = \sqrt{T/\mu} \), where \(\displaystyle T \) is the tension and \(\displaystyle \mu \) the linear mass density. Squaring and rearranging: \[T = \mu v^2 \] Using the unrounded speed \(\displaystyle v = 78.75\ \text{m s}^{-1} \): \[T = 4.0 \times 10^{-2}\ \text{kg m}^{-1} \times (78.75\ \text{m s}^{-1})^2 = 4.0 \times 10^{-2} \times 6201.5625\ \text{N} = 248.06\ \text{N} \]Every input in this problem ($\displaystyle 45$ Hz, \(\displaystyle 3.5\times10^{-2}\) kg, \(\displaystyle 4.0\times10^{-2}\) kg m\(\displaystyle ^{-1}\)) carries only two significant figures, so the final answers should not claim more precision than that. Rounding once, at the very end: \[T \approx 2.5 \times 10^{2}\ \text{N} \]A common slip here is mixing up mass and linear mass density, or using the rounded $\displaystyle 79$ m/s (instead of the exact $\displaystyle 78.75$ m/s) inside \(\displaystyle T = \mu v^2 \) — small rounding errors get squared and can shift the tension by several newtons.Answer: (a) speed of the transverse wave \(\displaystyle v \approx 79\ \text{m s}^{-1} \); (b) tension \(\displaystyle T \approx 2.5 \times 10^{2}\ \text{N} \) (≈ $\displaystyle 250$ N)

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 \(\displaystyle \lambda/4\) 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\[l_n=\frac{(2n-1)\lambda}{4},\qquad n=1,2,3,\dots \]which is the same statement as the hint \(\displaystyle v_n = \dfrac{4 l \nu}{2n-1}\). Here \(\displaystyle l\) is the length of air column, \(\displaystyle \lambda\) the wavelength of sound in the air, \(\displaystyle \nu\) the frequency and \(\displaystyle v\) the speed of sound.An aside on what is fixed and what is not: the frequency is set by the tuning fork, \(\displaystyle \nu = 340\ \text{Hz}\), 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 \(\displaystyle \nu\) is an input here, not something the tube decides.The first resonance. The shortest resonating length is the fundamental, \(\displaystyle n=1\):\[l_1=\frac{\lambda}{4}\quad\Longrightarrow\quad \lambda = 4l_1 = 4\times 0.255\ \text{m} = 1.02\ \text{m} \]The wave relation \(\displaystyle v=\nu\lambda\) then gives\[v = 340\ \text{s}^{-1}\times 1.02\ \text{m} = 346.8\ \text{m s}^{-1} \]Rounding once, at the end, to the three significant figures carried by both \(\displaystyle 340\ \text{Hz}\) and \(\displaystyle 25.5\ \text{cm}\):\[v \approx 347\ \text{m s}^{-1} \]What the second length is. With \(\displaystyle \lambda = 1.02\ \text{m}\), the next resonance is \(\displaystyle n=2\):\[l_2=\frac{3\lambda}{4}=\frac{3\times 1.02\ \text{m}}{4}=0.765\ \text{m}=76.5\ \text{cm} \]and the one after that would be \(\displaystyle 5\lambda/4 = 127.5\ \text{cm}\), 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 \(\displaystyle 79.3\ \text{cm}\) as measured, about \(\displaystyle 2.8\ \text{cm}\) longer than the ideal \(\displaystyle 76.5\ \text{cm}\). 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:\[\lambda = 2(l_2-l_1)=2(0.793-0.255)\ \text{m}=1.076\ \text{m},\qquad v=340\times1.076=366\ \text{m s}^{-1} \]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 \(\displaystyle \lambda=4l_1\) and \(\displaystyle \lambda=2(l_2-l_1)\) would have to hold, and they cannot both be true of these numbers — the data are over-determined and disagree by about \(\displaystyle 5\%\). Second, and decisively, check the two candidates against the temperature. In air, \(\displaystyle v \approx (331.3 + 0.606\,T)\ \text{m s}^{-1}\) with \(\displaystyle T\) in \(\displaystyle ^\circ\)C:\[347\ \text{m s}^{-1}\Rightarrow T\approx 26\,^\circ\text{C},\qquad 366\ \text{m s}^{-1}\Rightarrow T\approx 57\,^\circ\text{C} \]A room at \(\displaystyle 26\,^\circ\text{C}\) is an ordinary laboratory; \(\displaystyle 57\,^\circ\text{C}\) is not a temperature anyone runs a resonance-tube experiment at. The fundamental reading gives the physically sensible answer, and the \(\displaystyle 2.8\ \text{cm}\) surplus in the second reading is a blemish in the quoted data rather than a signal to switch methods.Answer: \(\displaystyle v = \nu\lambda = 340\ \text{Hz}\times 4(0.255\ \text{m}) = 346.8 \approx 347\ \text{m s}^{-1}\) (three significant figures), corresponding to an air temperature of about \(\displaystyle 26\,^\circ\text{C}\).

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 waveFor a standing wave, the distance between a node and the nearest antinode is always a quarter wavelength: \[\text{node-to-antinode distance} = \frac{\lambda}{4} \]The rod has length \(\displaystyle L = 100\ \text{cm} = 1.00\ \text{m}\), clamped exactly at its middle. So the node sits at the centre, and each free end is an antinode a distance \[l = \frac{L}{2} = \frac{1.00\ \text{m}}{2} = 0.50\ \text{m} \] away from that node. This half-length is what spans the quarter wavelength — not the full rod. (A common slip here is to set \(\displaystyle L = \lambda\) or \(\displaystyle L = \lambda/2\), 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\[\frac{\lambda}{4} = l = 0.50\ \text{m} \] \[\lambda = 4 \times 0.50\ \text{m} = 2.00\ \text{m} \]Finding the speedThe wave equation relates speed, frequency and wavelength: \[v = f\lambda \] where \(\displaystyle f\) is the fundamental frequency of longitudinal vibration, given as \(\displaystyle 2.53\ \text{kHz} = 2.53 \times 10^{3}\ \text{Hz}\).Substituting: \[v = (2.53 \times 10^{3}\ \text{Hz}) \times (2.00\ \text{m}) \] \[v = 5.06 \times 10^{3}\ \text{m/s} \]Both the given frequency ($\displaystyle 2.53$ kHz) and the given length ($\displaystyle 100$ cm, i.e. $\displaystyle 1.00$ m) carry three significant figures, so the answer is kept to three significant figures: \(\displaystyle 5.06 \times 10^{3}\ \text{m/s}\), i.e. \(\displaystyle 5.06\ \text{km/s}\). 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 ~$\displaystyle 340$ m/s speed of sound in air.Answer: The speed of sound (longitudinal wave) in steel is \(\displaystyle 5.06 \times 10^{3}\ \text{m/s} = 5.06\ \text{km/s}\).

A pipe closed at one end supports only the odd harmonics — find its fundamental frequency first, then see which multiple lands near $\displaystyle 430$ 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\[L = (2n-1)\frac{\lambda_n}{4}, \qquad n = 1, 2, 3, \dots \]so the resonant frequencies are\[f_n = (2n-1)\frac{v}{4L}, \qquad n = 1, 2, 3, \dots \]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 \(\displaystyle L = 20\text{ cm} = 0.20\text{ m}\) and the speed of sound is \(\displaystyle v = 340\text{ m s}^{-1}\).Fundamental (first harmonic), \(\displaystyle n=1\):\[f_1 = \frac{v}{4L} = \frac{340\text{ m s}^{-1}}{4 \times 0.20\text{ m}} = \frac{340}{0.80}\text{ Hz} = 425\text{ Hz} \]Third harmonic, \(\displaystyle n=2\):\[f_3 = 3f_1 = 3 \times 425\text{ Hz} = 1275\text{ Hz} \]The source frequency is $\displaystyle 430$ Hz. Compare it with the ladder of allowed frequencies: $\displaystyle 425$ Hz, $\displaystyle 1275$ Hz, $\displaystyle 2125$ Hz, … The third harmonic ($\displaystyle 1275$ Hz) and every harmonic above it are nowhere near $\displaystyle 430$ Hz, but the fundamental, $\displaystyle 425$ Hz, sits only $\displaystyle 5$ Hz away from the driving frequency — a difference of about $\displaystyle 1$%, well inside the rounding already present in the given data (\(\displaystyle L\) and \(\displaystyle v\) are given to only two significant figures). So the $\displaystyle 430$ 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:\[f_n' = n\frac{v}{2L}, \qquad n = 1, 2, 3, \dots \]Its fundamental is\[f_1' = \frac{v}{2L} = \frac{340\text{ m s}^{-1}}{2 \times 0.20\text{ m}} = \frac{340}{0.40}\text{ Hz} = 850\text{ Hz} \]and its allowed frequencies are $\displaystyle 850$ Hz, $\displaystyle 1700$ Hz, $\displaystyle 2550$ Hz, … A $\displaystyle 430$ Hz source is close to none of these — it is roughly half of $\displaystyle 850$ 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 $\displaystyle 430$ 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 \(\displaystyle v/4L\) to \(\displaystyle v/2L\) — it doubles — while also switching the harmonic content from odd multiples only to all integer multiples. Neither ladder happens to pass near $\displaystyle 430$ Hz once that end is opened.Answer: The $\displaystyle 430$ Hz source resonantly excites the first harmonic (fundamental, \(\displaystyle f_1 = 425\text{ Hz}\)) of the pipe closed at one end. With both ends open the fundamental becomes \(\displaystyle 850\text{ Hz}\) (and all higher harmonics are integer multiples of it), so the $\displaystyle 430$ Hz source would not be in resonance with the pipe in that case.

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 $\displaystyle 1$: Use the beat frequency to get the two possible values of B's frequency.When two sources of close frequencies \(\displaystyle f_A \) and \(\displaystyle f_B \) sound together, the beat frequency is \[f_{beat} = |f_A - f_B| \]Here \(\displaystyle f_A = 324 \) Hz and \(\displaystyle f_{beat} = 6 \) Hz, so \[|324 - f_B| = 6 \]This gives two candidates: \[f_B = 324 + 6 = 330 \ \text{Hz} \qquad \text{or} \qquad f_B = 324 - 6 = 318 \ \text{Hz} \]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 $\displaystyle 2$: Relate tension to frequency for a stretched string.For a string under tension \(\displaystyle T \), the fundamental frequency is \[f \propto \sqrt{T} \]So reducing the tension in A lowers its frequency — \(\displaystyle f_A \) moves down from $\displaystyle 324$ Hz to some new value \(\displaystyle f_A' < 324 \) 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 $\displaystyle 3$: Test each candidate against the new (smaller) beat frequency.If \(\displaystyle f_B = 330 \) Hz (B above A): lowering \(\displaystyle f_A \) below $\displaystyle 324$ Hz pulls A further away from $\displaystyle 330$ Hz, so the gap \(\displaystyle |f_A' - 330| \) would grow larger than $\displaystyle 6$ Hz — the beat frequency would increase. This contradicts the given fact that the beats dropped to $\displaystyle 3$ Hz, so this value of \(\displaystyle f_B \) is ruled out.If \(\displaystyle f_B = 318 \) Hz (B below A): lowering \(\displaystyle f_A \) from $\displaystyle 324$ Hz moves A closer to $\displaystyle 318$ Hz, so the gap \(\displaystyle |f_A' - 318| \) shrinks below $\displaystyle 6$ Hz — the beat frequency decreases, exactly as observed (it falls to $\displaystyle 3$ Hz). This is consistent.Step $\displaystyle 4$: Conclusion.Only \(\displaystyle f_B = 318 \) Hz is consistent with the beat frequency falling from $\displaystyle 6$ Hz to $\displaystyle 3$ Hz when A's tension — and hence A's frequency — is reduced.Answer: The frequency of string B is $\displaystyle 318$ Hz.

(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 \(\displaystyle s(x,t)\) through Hooke's law for a volume element, using the bulk modulus \(\displaystyle B\): \[p(x,t) = -B\,\frac{\partial s}{\partial x} \] For a standing sound wave (as in an air column) the displacement pattern is \[s(x,t) = s_0 \sin(kx)\cos(\omega t) \] Differentiating with respect to \(\displaystyle x\) gives the pressure variation \[p(x,t) = -B s_0 k \cos(kx)\cos(\omega t) \] A displacement node is where \(\displaystyle \sin(kx) = 0\), so \(\displaystyle s = 0\) there for all time — but at exactly that \(\displaystyle x\), \(\displaystyle \cos(kx) = \pm 1\), 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 \(\displaystyle \sin(kx) = \pm 1\) (maximum swing of the air layer), but there \(\displaystyle \cos(kx) = 0\), so \(\displaystyle p = 0\) 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 $\displaystyle 100$ kHz — well above the human hearing limit of about $\displaystyle 20$ kHz. Using \(\displaystyle v = f\lambda\) with the speed of sound in air \(\displaystyle v \approx 340\ \text{m/s}\), 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 \(\displaystyle \lambda \ll\) size of the obstacle). This lets the returning echo carry sharp, direction-specific information rather than being smeared out. \[d = \frac{v\,\Delta t}{2} \] 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: \[s(x,t) = \sum_n A_n \sin(k_n x)\cos(\omega_n t + \phi_n) \] If two instruments are tuned so their fundamental frequency \(\displaystyle \omega_1\) is the same, the two notes have the same pitch. But the relative amplitudes \(\displaystyle A_n\) and phases \(\displaystyle \phi_n\) 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. 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 \(\displaystyle B \neq 0\). 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): \[s(x,t) = \int A(k)\cos(kx-\omega(k)\,t)\,dk \] In a non-dispersive medium, every one of these component waves travels at the same phase speed \(\displaystyle v = \omega/k\) = constant, independent of \(\displaystyle k\). Their relative phase relationship never changes as they propagate, so the sum of all the components — the pulse — simply translates along at speed \(\displaystyle v\) without changing shape.In a dispersive medium, the relationship between \(\displaystyle \omega\) and \(\displaystyle k\) is not a straight line through the origin, so \(\displaystyle v = \omega/k\) 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 \(\displaystyle v_g = d\omega/dk\), 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.