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401. For a right-moving wave \(y=F(x-vt)\), the relation between particle velocity and wave-profile slope is
ⓐ. \(\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}\)
ⓑ. \(\frac{\partial y}{\partial t}=v\frac{\partial y}{\partial x}\)
ⓒ. \(\frac{\partial y}{\partial t}=\frac{1}{v}\frac{\partial y}{\partial x}\)
ⓓ. \(\frac{\partial y}{\partial t}=v^2\frac{\partial y}{\partial x}\)
Correct Answer: \(\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}\)
Explanation: \( \textbf{Given wave form:} \)
\[
y=F(x-vt)
\]
\( \textbf{Let the argument be:} \)
\[
u=x-vt
\]
\( \textbf{Spatial derivative:} \)
\[
\frac{\partial y}{\partial x}=F'(u)\frac{\partial u}{\partial x}=F'(u)
\]
\( \textbf{Time derivative:} \)
\[
\frac{\partial y}{\partial t}=F'(u)\frac{\partial u}{\partial t}=F'(u)(-v)
\]
\( \textbf{Combining both results:} \)
\[
\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}
\]
\( \textbf{Final answer:} \) \(\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}\).
The negative sign is tied to rightward translation of the profile, not to a negative wave speed.
402. A left-moving wave is described by \(y=G(x+vt)\). If a point on the graph has positive slope \(\frac{\partial y}{\partial x}\gt0\), the particle velocity \(\frac{\partial y}{\partial t}\) at that point is
ⓐ. negative
ⓑ. zero for every left-moving wave
ⓒ. unrelated to the slope
ⓓ. positive
Correct Answer: positive
Explanation: For a left-moving wave, the displacement may be written as \(y=G(x+vt)\). Let \(u=x+vt\), so \(y=G(u)\). The spatial slope is \(\frac{\partial y}{\partial x}=G'(u)\). The time derivative is \(\frac{\partial y}{\partial t}=vG'(u)\). Therefore, \(\frac{\partial y}{\partial t}=v\frac{\partial y}{\partial x}\) for a left-moving wave. Since \(v\) is positive and the slope is positive, the particle velocity is also positive. The sign relation is opposite to that for a right-moving profile.
403. A sinusoidal travelling wave \(y=A\sin(kx-\omega t)\) is tested in the one-dimensional wave equation. The suitable wave equation and speed relation are
ⓐ. \(\frac{\partial^2 y}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}\), with \(v=\frac{\omega}{k}\)
ⓑ. \(\frac{\partial^2 y}{\partial t^2}=\frac{1}{v^2}\frac{\partial^2 y}{\partial x^2}\), with \(v=\frac{k}{\omega}\)
ⓒ. \(\frac{\partial y}{\partial x}=\frac{1}{v^2}\frac{\partial y}{\partial t}\), with \(v=\omega k\)
ⓓ. \(\frac{\partial^2 y}{\partial x^2}=v^2\frac{\partial^2 y}{\partial t^2}\), with \(v=\frac{k}{\omega}\)
Correct Answer: \(\frac{\partial^2 y}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}\), with \(v=\frac{\omega}{k}\)
Explanation: \( \textbf{Given wave:} \)
\[
y=A\sin(kx-\omega t)
\]
\( \textbf{Second derivative with respect to \(x\):} \)
\[
\frac{\partial^2 y}{\partial x^2}=-k^2y
\]
\( \textbf{Second derivative with respect to \(t\):} \)
\[
\frac{\partial^2 y}{\partial t^2}=-\omega^2y
\]
\( \textbf{Compare the two:} \)
\[
\frac{\partial^2 y}{\partial x^2}=\frac{k^2}{\omega^2}\frac{\partial^2 y}{\partial t^2}
\]
\( \textbf{Using \(v=\frac{\omega}{k}\):} \)
\[
\frac{k^2}{\omega^2}=\frac{1}{v^2}
\]
\( \textbf{Final answer:} \) The wave satisfies \(\frac{\partial^2 y}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}\), with \(v=\frac{\omega}{k}\).
The speed relation comes from matching the spatial and time curvatures of the same travelling pattern.
404. For a travelling wave on a string, the transverse force component is approximately related to the slope by \(F_y=T_s\frac{\partial y}{\partial x}\) for small slopes. The average power carried by a sinusoidal wave is proportional to
ⓐ. \(\frac{A}{\omega v}\)
ⓑ. \(A\omega k^0\) only
ⓒ. \(\frac{v}{A^2\omega^2}\)
ⓓ. \(A^2\omega^2v\)
Correct Answer: \(A^2\omega^2v\)
Explanation: The average power carried by a sinusoidal wave on a string is commonly written as \(P_{\text{avg}}=\frac{1}{2}\mu\omega^2A^2v\). This relation shows that power depends on the square of amplitude and the square of angular frequency. It also depends directly on wave speed \(v\) and linear mass density \(\mu\). Therefore, keeping the string and medium conditions fixed, the wave power is proportional to \(A^2\omega^2v\). A direct dependence on \(A\) alone would miss the energy nature of oscillation. The square dependence is why even moderate changes in amplitude or frequency can strongly change transmitted power.
405. A sinusoidal wave travels on the same stretched string. Its amplitude is doubled and its frequency is also doubled, while the string tension and linear mass density remain unchanged. The average power carried by the wave becomes
ⓐ. \(4\) times
ⓑ. \(8\) times
ⓒ. \(16\) times
ⓓ. \(32\) times
Correct Answer: \(16\) times
Explanation: \( \textbf{Power dependence for a sinusoidal string wave:} \)
\[
P_{\text{avg}}\propto A^2\omega^2v
\]
\( \textbf{Same string condition:} \) The wave speed \(v\) remains unchanged because \(T_s\) and \(\mu\) are unchanged.
\( \textbf{Amplitude change:} \)
\[
A'=2A
\]
\( \textbf{Frequency and angular frequency change:} \)
\[
f'=2f,\quad \omega'=2\omega
\]
\( \textbf{Power ratio:} \)
\[
\frac{P'_{\text{avg}}}{P_{\text{avg}}}=\left(\frac{A'}{A}\right)^2\left(\frac{\omega'}{\omega}\right)^2
\]
\( \textbf{Substitution:} \)
\[
\frac{P'_{\text{avg}}}{P_{\text{avg}}}=2^2\times2^2
\]
\( \textbf{Calculation:} \)
\[
\frac{P'_{\text{avg}}}{P_{\text{avg}}}=16
\]
\( \textbf{Final answer:} \) The average power becomes \(16\) times.
The speed does not add another factor because the same string under the same tension carries the disturbance with the same wave speed.
406. A wave on a string has \(A=0.010\,\text{m}\), \(\omega=100\,\text{rad s}^{-1}\), \(\mu=0.020\,\text{kg m}^{-1}\), and \(v=50\,\text{m s}^{-1}\). Using \(P_{\text{avg}}=\frac{1}{2}\mu\omega^2A^2v\), the average power is
ⓐ. \(0.25\,\text{W}\)
ⓑ. \(0.50\,\text{W}\)
ⓒ. \(2.5\,\text{W}\)
ⓓ. \(5.0\,\text{W}\)
Correct Answer: \(0.50\,\text{W}\)
Explanation: \( \textbf{Given data:} \) \(A=0.010\,\text{m}\), \(\omega=100\,\text{rad s}^{-1}\), \(\mu=0.020\,\text{kg m}^{-1}\), \(v=50\,\text{m s}^{-1}\).
\( \textbf{Required quantity:} \) Average power \(P_{\text{avg}}\).
\( \textbf{Formula:} \)
\[
P_{\text{avg}}=\frac{1}{2}\mu\omega^2A^2v
\]
\( \textbf{Substitution:} \)
\[
P_{\text{avg}}=\frac{1}{2}(0.020)(100)^2(0.010)^2(50)
\]
\( \textbf{Evaluate factors:} \)
\[
(100)^2=10000,\quad (0.010)^2=0.0001
\]
\( \textbf{Product inside formula:} \)
\[
10000\times0.0001=1
\]
\( \textbf{Remaining calculation:} \)
\[
P_{\text{avg}}=\frac{1}{2}(0.020)(1)(50)
\]
\[
P_{\text{avg}}=0.50\,\text{W}
\]
\( \textbf{Final answer:} \) The average power is \(0.50\,\text{W}\).
The amplitude must be squared, so using \(0.010\) instead of \(0.0001\) would overestimate the power.
407. A transverse sinusoidal wave on a string has displacement \(y=A\sin(kx-\omega t)\). At a point where \(y=0\), the particle acceleration and particle speed are respectively
ⓐ. zero and maximum
ⓑ. maximum and zero
ⓒ. maximum and maximum
ⓓ. zero and zero
Correct Answer: zero and maximum
Explanation: A particle of the string executes simple harmonic motion while the wave passes. For the sinusoidal wave, particle acceleration is \(a_y=-\omega^2y\). At the mean position \(y=0\), this gives \(a_y=0\). The particle speed is maximum at the mean position because the particle is passing through equilibrium. This is the same phase relation as ordinary simple harmonic motion. The condition is about local particle motion, not about the speed of propagation of the wave pattern. The wave can still be travelling even when a particular particle has zero acceleration at that instant.
408. A point on a string has zero transverse velocity at a certain instant in a sinusoidal travelling wave. Its displacement is most likely
ⓐ. zero
ⓑ. at an extreme value
ⓒ. equal to one wavelength
ⓓ. equal to the wave speed
Correct Answer: at an extreme value
Explanation: In a sinusoidal wave, each string particle performs simple harmonic motion. The transverse particle velocity becomes zero at the extreme positions of the oscillation. These positions correspond to maximum positive displacement \(+A\) or maximum negative displacement \(-A\). At the mean position, the transverse particle speed is maximum, not zero. Wavelength is a distance between phase-matching points along the string and is not a possible transverse displacement of one particle. The zero-velocity condition identifies the turning points of the particle’s motion.
409. A graph of displacement \(y\) versus position \(x\) for a right-moving wave has a negative slope at point P. At that instant, the transverse velocity of the particle at P is
ⓐ. negative
ⓑ. zero only because the slope is non-zero
ⓒ. equal to the wave speed along \(x\)
ⓓ. positive
Correct Answer: positive
Explanation: For a right-moving wave \(y=F(x-vt)\), the particle velocity and spatial slope are related by
\[
\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}
\]
Here \(v\) is positive. If the graph slope \(\frac{\partial y}{\partial x}\) is negative, then \(-v\frac{\partial y}{\partial x}\) is positive. Therefore, the particle is moving upward if positive \(y\) is upward. This result uses the motion of the travelling profile, not the local height alone. A non-zero slope does not mean the particle velocity must equal the wave speed, because the two describe different motions.
410. A sinusoidal wave is written as \(y=0.030\sin(2\pi x-10\pi t)\), with SI units. At \(x=0\) and \(t=0\), the particle velocity is
ⓐ. \(+0.30\pi\,\text{m s}^{-1}\)
ⓑ. \(-0.30\pi\,\text{m s}^{-1}\)
ⓒ. \(0\,\text{m s}^{-1}\)
ⓓ. \(10\pi\,\text{m s}^{-1}\)
Correct Answer: \(-0.30\pi\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given wave:} \)
\[
y=0.030\sin(2\pi x-10\pi t)
\]
\( \textbf{Compare with standard form:} \)
\[
y=A\sin(kx-\omega t)
\]
\( \textbf{Identify constants:} \)
\[
A=0.030\,\text{m},\quad \omega=10\pi\,\text{s}^{-1}
\]
\( \textbf{Particle velocity:} \)
\[
v_y=\frac{\partial y}{\partial t}=-A\omega\cos(kx-\omega t)
\]
\( \textbf{At \(x=0,t=0\):} \)
\[
kx-\omega t=0
\]
\[
\cos0=1
\]
\( \textbf{Substitution:} \)
\[
v_y=-(0.030)(10\pi)(1)
\]
\( \textbf{Final answer:} \) \(v_y=-0.30\pi\,\text{m s}^{-1}\).
The displacement is zero at that instant, but the particle velocity is not zero because the particle is crossing its mean position.
411. A standing wave on a string is represented by \(y=2A\sin kx\cos\omega t\). The positions of nodes satisfy
ⓐ. \(\sin kx=0\)
ⓑ. \(\cos\omega t=0\)
ⓒ. \(\sin kx=1\)
ⓓ. \(\cos\omega t=1\)
Correct Answer: \(\sin kx=0\)
Explanation: In a standing wave, a node is a position where displacement remains zero at all times. The expression \(y=2A\sin kx\cos\omega t\) separates the position factor and the time factor. For a point to be a node for every value of time, the position factor must be zero. Therefore, \(\sin kx=0\) gives the node positions. Values of \(\cos\omega t=0\) make the whole string momentarily pass through zero displacement, but that does not define fixed nodes. A node is a permanent position of zero amplitude, not just a zero displacement at one instant.
412. For the standing wave \(y=2A\sin kx\cos\omega t\), an antinode occurs where
ⓐ. \(\sin kx=0\)
ⓑ. \(\cos\omega t=0\) only
ⓒ. \(kx=\omega t\) only
ⓓ. \(|\sin kx|=1\)
Correct Answer: \(|\sin kx|=1\)
Explanation: The amplitude at position \(x\) in the standing wave is controlled by the factor \(2A|\sin kx|\). Antinodes are points where this amplitude is maximum. The maximum value of \(|\sin kx|\) is \(1\), so antinodes satisfy \(|\sin kx|=1\). Nodes instead satisfy \(\sin kx=0\). The factor \(\cos\omega t\) changes the displacement of all points with time, but it does not decide the fixed antinode positions. Antinode position is a spatial condition, not a condition at one particular instant.
413. Two waves \(y_1=A\sin(kx-\omega t)\) and \(y_2=A\sin(kx+\omega t)\) overlap on a string. The resultant wave is
ⓐ. \(y=2A\sin kx\cos\omega t\)
ⓑ. \(y=2A\cos kx\sin\omega t\)
ⓒ. \(y=A\sin(2kx)\)
ⓓ. \(y=A\sin(2\omega t)\)
Correct Answer: \(y=2A\sin kx\cos\omega t\)
Explanation: \( \textbf{Given waves:} \)
\[
y_1=A\sin(kx-\omega t),\quad y_2=A\sin(kx+\omega t)
\]
\( \textbf{Superposition:} \)
\[
y=y_1+y_2
\]
\( \textbf{Identity used:} \)
\[
\sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)
\]
\( \textbf{Here:} \)
\[
C=kx-\omega t,\quad D=kx+\omega t
\]
\( \textbf{Average of phases:} \)
\[
\frac{C+D}{2}=kx
\]
\( \textbf{Half-difference:} \)
\[
\frac{C-D}{2}=-\omega t
\]
\( \textbf{Since \(\cos(-\omega t)=\cos\omega t\):} \)
\[
y=2A\sin kx\cos\omega t
\]
\( \textbf{Final answer:} \) The resultant is \(y=2A\sin kx\cos\omega t\).
The separate \(x\)-factor and \(t\)-factor show that the result is a standing wave.
414. In a standing wave, all particles between two neighbouring nodes
ⓐ. oscillate with the same amplitude but random phases
ⓑ. remain permanently at rest
ⓒ. move along the string from one node to the next
ⓓ. oscillate in phase with different amplitudes
Correct Answer: oscillate in phase with different amplitudes
Explanation: In one loop of a standing wave, all particles between two adjacent nodes pass through their mean positions together and reach extreme positions together. They therefore oscillate in phase within that loop. Their amplitudes are not the same; amplitude is zero at the nodes and maximum at the antinode. In the neighbouring loop, the particles oscillate in opposite phase. The particles do not travel along the string from one node to another. Standing waves show local oscillation with fixed node and antinode positions.
415. Two points in adjacent loops of a standing wave on a string have their displacements at the same instant compared. If one loop is displaced upward, the adjacent loop is
ⓐ. also displaced upward
ⓑ. displaced downward
ⓒ. always at rest
ⓓ. moving along the string
Correct Answer: displaced downward
Explanation: Adjacent loops in a standing wave are separated by a node. The displacement pattern changes sign when crossing a node. Therefore, particles in neighbouring loops oscillate in opposite phase. If one loop is displaced upward at a given instant, the adjacent loop is displaced downward. Both loops pass through the mean position at the same times, but their extreme displacements are opposite. This is a phase property of the standing-wave pattern, not a sign that energy is being transported from loop to loop.
416. A standing wave is observed on a string. The distance between the first and the fifth node is \(0.80\,\text{m}\). The wavelength is
ⓐ. \(0.20\,\text{m}\)
ⓑ. \(0.80\,\text{m}\)
ⓒ. \(0.40\,\text{m}\)
ⓓ. \(1.60\,\text{m}\)
Correct Answer: \(0.40\,\text{m}\)
Explanation: \( \textbf{Given information:} \) Distance from the first node to the fifth node is \(0.80\,\text{m}\).
\( \textbf{Node spacing rule:} \)
\[
\text{distance between adjacent nodes}=\frac{\lambda}{2}
\]
\( \textbf{Number of intervals:} \) From the first node to the fifth node, there are \(4\) adjacent node intervals.
\( \textbf{So:} \)
\[
4\left(\frac{\lambda}{2}\right)=0.80\,\text{m}
\]
\( \textbf{Simplifying:} \)
\[
2\lambda=0.80\,\text{m}
\]
\( \textbf{Solving:} \)
\[
\lambda=0.40\,\text{m}
\]
\( \textbf{Final answer:} \) The wavelength is \(0.40\,\text{m}\).
Counting intervals rather than counting nodes is the key step in this spacing problem.
417. A string fixed at both ends vibrates in a mode whose neighbouring nodes are \(0.25\,\text{m}\) apart. If the wave speed on the string is \(100\,\text{m s}^{-1}\), the frequency of the wave is
ⓐ. \(100\,\text{Hz}\)
ⓑ. \(200\,\text{Hz}\)
ⓒ. \(400\,\text{Hz}\)
ⓓ. \(800\,\text{Hz}\)
Correct Answer: \(200\,\text{Hz}\)
Explanation: \( \textbf{Given node spacing:} \)
\[
\frac{\lambda}{2}=0.25\,\text{m}
\]
\( \textbf{Find wavelength:} \)
\[
\lambda=0.50\,\text{m}
\]
\( \textbf{Given wave speed:} \)
\[
v=100\,\text{m s}^{-1}
\]
\( \textbf{Wave relation:} \)
\[
v=f\lambda
\]
\( \textbf{Rearranging:} \)
\[
f=\frac{v}{\lambda}
\]
\( \textbf{Substitution:} \)
\[
f=\frac{100\,\text{m s}^{-1}}{0.50\,\text{m}}
\]
\( \textbf{Calculation:} \)
\[
f=200\,\text{Hz}
\]
\( \textbf{Final answer:} \) The frequency is \(200\,\text{Hz}\).
The given spacing is half a wavelength, not the full wavelength.
418. In an open pipe, end correction is introduced because
ⓐ. the antinode lies slightly beyond the open end
ⓑ. the closed end becomes a displacement antinode
ⓒ. sound stops exactly at the geometrical end
ⓓ. the air column no longer has pressure variation
Correct Answer: the antinode lies slightly beyond the open end
Explanation: At an open end, air just outside the pipe also participates in the oscillation. As a result, the displacement antinode is not exactly at the geometrical end of the pipe. It lies slightly outside the open end. End correction accounts for this extra effective length. This is why an open pipe has effective length larger than its physical length. The correction changes resonance calculations without changing the basic boundary condition that an open end is a displacement antinode and pressure node.
419. A closed pipe has physical length \(0.32\,\text{m}\) and end correction \(0.02\,\text{m}\) at its open end. If the speed of sound is \(340\,\text{m s}^{-1}\), the fundamental frequency is closest to
ⓐ. \(265\,\text{Hz}\)
ⓑ. \(531\,\text{Hz}\)
ⓒ. \(1063\,\text{Hz}\)
ⓓ. \(250\,\text{Hz}\)
Correct Answer: \(250\,\text{Hz}\)
Explanation: \( \textbf{Given data:} \) Physical length \(L=0.32\,\text{m}\), end correction \(e=0.02\,\text{m}\), sound speed \(v=340\,\text{m s}^{-1}\).
\( \textbf{Effective length for a closed-open pipe:} \)
\[
L_{\text{eff}}=L+e
\]
\( \textbf{Substitution:} \)
\[
L_{\text{eff}}=0.32\,\text{m}+0.02\,\text{m}=0.34\,\text{m}
\]
\( \textbf{Fundamental condition:} \)
\[
L_{\text{eff}}=\frac{\lambda}{4}
\]
\( \textbf{Frequency relation:} \)
\[
f_1=\frac{v}{4L_{\text{eff}}}
\]
\( \textbf{Substitution:} \)
\[
f_1=\frac{340}{4(0.34)}
\]
\( \textbf{Denominator:} \)
\[
4(0.34)=1.36
\]
\( \textbf{Calculation:} \)
\[
f_1=250\,\text{Hz}
\]
\( \textbf{Final answer:} \) The fundamental frequency is closest to \(250\,\text{Hz}\).
Only one end correction is added because only one end of the pipe is open.
420. In a resonance tube closed by water, the first two resonance lengths are \(L_1\) and \(L_2\). If end correction is present but unchanged, the difference \(L_2-L_1\) equals
ⓐ. \(\frac{\lambda}{4}\)
ⓑ. \(\frac{\lambda}{2}\)
ⓒ. \(\lambda\)
ⓓ. \(2\lambda\)
Correct Answer: \(\frac{\lambda}{2}\)
Explanation: For a closed-open resonance tube, the effective resonance lengths are approximately \(\frac{\lambda}{4}\), \(\frac{3\lambda}{4}\), \(\frac{5\lambda}{4}\), and so on. If end correction \(e\) is present, the effective length is \(L+e\). The same \(e\) is added to each resonance length for the same tube opening. Therefore, when subtracting two successive physical resonance lengths, the end correction cancels. The separation becomes \(\frac{3\lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2}\). This is why successive resonance differences are useful for finding wavelength even when end correction is not known.