Oscillations and Waves MCQs with Answers – Part 5 (NEET Physics)

Explanation: \(\textbf{Model for the recorded timings:}\) The true duration of \(N\) complete oscillations is \(NT\). The instrument adds the same delay \(\delta\) to each reading. Therefore, \[ \tau=NT+\delta \] A graph of recorded time against oscillation count would have slope \(T\) and intercept \(\delta\). \(\textbf{Eliminating the systematic delay:}\) Using the readings for 20 and 50 oscillations, \[ 100.30-40.30=(50-20)T \] \[ 60.00=30T \] \[ T=2.000\,\text{s} \] \(\textbf{Recovering the fixed delay:}\) Substitute the period into the first timing equation: \[ 40.30=20(2.000)+\delta \] \[ \delta=0.30\,\text{s} \] The third reading independently gives \[ 80(2.000)+0.30=160.30\,\text{s} \] so the observations are consistent. \(\textbf{Determining gravitational acceleration:}\) For an ideal small-angle pendulum, \[ T=2\pi\sqrt{\frac Lg} \] Therefore, \[ g=\frac{4\pi^2L}{T^2} \] Substituting, \[ g=\frac{4\pi^2(1.00)}{(2.000)^2} \] \[ g=\pi^2\,\text{m s}^{-2} \] \(\textbf{Error from ignoring the delay:}\) The naive period obtained from the first reading would be \[ T_{\mathrm{naive}}=\frac{40.30}{20} \] \[ T_{\mathrm{naive}}=2.015\,\text{s} \] Its absolute error is \[ \Delta T=2.015-2.000 \] \[ \Delta T=0.015\,\text{s} \] Thus, \[ \text{Percentage error}=\frac{0.015}{2.000}\times100 \] \[ \text{Percentage error}=0.75\% \] \(\textbf{Physical interpretation:}\) A fixed additive timing delay is a systematic error, not an uncertainty that necessarily disappears when the number of oscillations increases. However, taking the difference between two recorded durations removes that common delay. The true pendulum period is therefore obtained from the change in recorded time per additional oscillation. The third measurement verifies the linear timing model, and the effective length then gives gravity. \(\textbf{Final answer:}\) \[ \boxed{T=2.000\,\text{s},\qquad \delta=0.30\,\text{s},\qquad g=\pi^2\,\text{m s}^{-2},\qquad \text{period error}=0.75\%} \]

Explanation: \(\textbf{Effective gravity in the accelerating lift:}\) For upward lift acceleration \(a\), the effective gravitational acceleration is \[ g_{\mathrm{eff}}=g+a \] The small-angle pendulum period is \[ T=2\pi\sqrt{\frac{L}{g+a}} \] \(\textbf{Linearising the period relation:}\) Squaring and taking the reciprocal, \[ \frac1{T^2}=\frac{g+a}{4\pi^2L} \] Therefore, \[ \frac1{T^2}=\frac{1}{4\pi^2L}a+\frac{g}{4\pi^2L} \] The graph slope determines \(1/(4\pi^2L)\), while its vertical intercept determines \(g/(4\pi^2L)\). \(\textbf{Slope from the measured points:}\) \[ S=\frac{1-1/4}{3\pi^2-0} \] \[ S=\frac{3/4}{3\pi^2} \] \[ S=\frac1{4\pi^2} \] Therefore, \[ \frac1{4\pi^2L}=\frac1{4\pi^2} \] \[ L=1.00\,\text{m} \] \(\textbf{Recovering local gravity from the intercept:}\) At \(a=0\), \[ \frac1{T_0^2}=\frac14\,\text{s}^{-2} \] Thus, \[ \frac{g}{4\pi^2L}=\frac14 \] Substituting \(L=1.00\,\text{m}\), \[ g=\pi^2\,\text{m s}^{-2} \] \(\textbf{Period during downward acceleration:}\) For downward acceleration of magnitude \(g/2\), \[ a=-\frac g2 \] Hence, \[ g_{\mathrm{eff}}=g-\frac g2=\frac g2 \] The required period is \[ T'=2\pi\sqrt{\frac{L}{g/2}} \] Substituting, \[ T'=2\pi\sqrt{\frac{2(1.00)}{\pi^2}} \] \[ T'=2\sqrt2\,\text{s} \] \(\textbf{Physical interpretation:}\) Plotting reciprocal squared period against lift acceleration produces a straight line, whereas period itself is not linear in acceleration. The slope and intercept independently determine pendulum length and gravitational acceleration. Downward lift acceleration weakens effective gravity, increasing the period. The graph also reveals that the period formula ceases to describe oscillatory motion when the effective gravitational acceleration becomes zero. \(\textbf{Final answer:}\) \[ \boxed{L=1.00\,\text{m},\qquad g=\pi^2\,\text{m s}^{-2},\qquad T'=2\sqrt2\,\text{s}} \]

Explanation: \(\textbf{Tension at the release extreme:}\) At the extreme, the bob is momentarily at rest. Its radial acceleration is therefore zero. Resolving the weight along the string, \[ S_{\mathrm{ext}}=mg\cos\Theta \] Using the measured tension, \[ 3.0=0.50g\cos\Theta \] Therefore, \[ g\cos\Theta=6.0\,\text{m s}^{-2} \] \(\textbf{Energy gained on descending to equilibrium:}\) The bob descends through height \[ h=L(1-\cos\Theta) \] By conservation of mechanical energy, \[ \frac12mv_{\max}^2=mgL(1-\cos\Theta) \] Hence, \[ v_{\max}^2=2gL(1-\cos\Theta) \] \(\textbf{Radial equation at the lowest position:}\) At equilibrium, string tension exceeds weight by the required centripetal force. \[ S_{\mathrm{bottom}}-mg=\frac{mv_{\max}^2}{L} \] Substituting the energy-derived speed, \[ S_{\mathrm{bottom}}=mg+2mg(1-\cos\Theta) \] \[ S_{\mathrm{bottom}}=mg(3-2\cos\Theta) \] \(\textbf{Eliminating the unknown release angle:}\) The measured bottom tension is \(9.0\,\text{N}\). \[ 9.0=0.50g(3-2\cos\Theta) \] Using \(0.50g\cos\Theta=3.0\), \[ 9.0=1.50g-6.0 \] \[ 1.50g=15.0 \] \[ g=10\,\text{m s}^{-2} \] \(\textbf{Recovering the release angle:}\) From the extreme-position tension, \[ 3.0=0.50(10)\cos\Theta \] \[ \cos\Theta=\frac35 \] Therefore, \[ \Theta=\cos^{-1}\left(\frac35\right) \] \(\textbf{Maximum bob speed:}\) \[ v_{\max}^2=2gL(1-\cos\Theta) \] Substituting, \[ v_{\max}^2=2(10)(1.00)\left(1-\frac35\right) \] \[ v_{\max}^2=8.0\,\text{m}^2\text{s}^{-2} \] \[ v_{\max}=2\sqrt2\,\text{m s}^{-1} \] \(\textbf{Physical interpretation:}\) The tension at an extreme depends on the radial component of gravitational force because the speed vanishes there. At the lowest point, additional tension supplies centripetal acceleration. The difference between the two readings therefore contains information about both gravitational acceleration and the release angle. Energy conservation connects these readings without requiring a small-angle approximation. The inferred speed and angle also satisfy the condition that the string remain taut throughout the swing. \(\textbf{Final answer:}\) \[ \boxed{g=10\,\text{m s}^{-2},\qquad \Theta=\cos^{-1}\left(\frac35\right),\qquad v_{\max}=2\sqrt2\,\text{m s}^{-1}} \]

Explanation: Both pendulums have the same length and gravitational acceleration, so they have the same small-angle angular frequency. Their different amplitudes do not change that frequency within the stipulated approximation. For P, the positive displacement \(+\Theta/2\) combined with positive angular velocity corresponds to a cosine phase of \(5\pi/3\). For Q, displacement \(-\Theta\) relative to angular amplitude \(2\Theta\), together with positive angular velocity, corresponds to a cosine phase of \(4\pi/3\). Therefore, the two phases differ by a constant amount of \(\pi/3\), modulo \(2\pi\). A pendulum crosses equilibrium whenever its cosine phase is \(\pi/2\) or \(3\pi/2\), modulo \(2\pi\). If both pendulums crossed equilibrium simultaneously, their phases at that instant would differ by an integer multiple of \(\pi\). Their actual phase difference is \(\pi/3\), which is not an integer multiple of \(\pi\). Since both phases advance at the same angular frequency, this difference never changes. Thus, simultaneous equilibrium passage does not occur. The fact that both are initially moving towards positive displacement is not sufficient to synchronise their later equilibrium crossings.

Explanation: \(\textbf{Translating the two pulse intervals:}\) Pulse P moves rightwards by \(vt\). Its occupied interval at time \(t\) is \[ 2t\lt x\lt 1+2t \] Pulse Q moves leftwards by the same distance. Its occupied interval is \[ 3-2t\lt x\lt 4-2t \] \(\textbf{Pulse positions at the observation time:}\) At \[ t=0.75\,\text{s} \] the translation distance is \[ vt=2.0(0.75)=1.50\,\text{m} \] Thus, P occupies \[ 1.50\lt x\lt 2.50 \] and Q also occupies \[ 1.50\lt x\lt 2.50 \] The two rectangular pulses completely overlap. \(\textbf{Resultant displacement during overlap:}\) For an ideal linear string, the transverse displacements add algebraically. \[ y_{\mathrm{resultant}}=y_P+y_Q \] Substituting signed pulse heights, \[ y_{\mathrm{resultant}}=+3.0-2.0 \] \[ y_{\mathrm{resultant}}=+1.0\,\text{cm} \] The upward pulse is not destroyed by the downward pulse. \(\textbf{Determining when overlap ends:}\) After complete overlap, P continues rightwards and Q continues leftwards. The last common point disappears when P's left edge reaches Q's right edge. \[ 2t=4-2t \] \[ 4t=4 \] \[ t=1.00\,\text{s} \] \(\textbf{Checking the separation afterwards:}\) At any time greater than \(1.00\,\text{s}\), P's left edge lies to the right of Q's right edge. The pulses are then separated and continue travelling with their original amplitudes and shapes. \(\textbf{Physical interpretation:}\) Complete overlap produces the algebraic sum of the signed displacements. Because the pulse heights are unequal, destructive superposition leaves a residual upward displacement rather than complete cancellation. The pulses subsequently separate without permanent loss of their individual shapes in the stipulated linear medium. The end of overlap is determined geometrically from the moving pulse boundaries, not from the instant of maximum cancellation. \(\textbf{Final answer:}\) \[ \boxed{1.50\lt x\lt 2.50\,\text{m},\qquad y_{\mathrm{resultant}}=+1.0\,\text{cm},\qquad t_{\mathrm{end}}=1.00\,\text{s}} \]

Explanation: \(\textbf{Superposing waves with a common travelling phase:}\) Write \[ \phi=kx-\omega t \] The resultant displacement is the algebraic sum: \[ y=y_1+y_2 \] \[ y=3\sin\phi+4\cos\phi\quad\text{mm} \] \(\textbf{Converting the sum into one harmonic wave:}\) Write the resultant as \[ y=R\cos(\phi-\delta) \] Expanding, \[ y=R\cos\delta\cos\phi+R\sin\delta\sin\phi \] Comparison of coefficients gives \[ R\cos\delta=4,\qquad R\sin\delta=3 \] Therefore, \[ R=\sqrt{4^2+3^2} \] \[ R=5\,\text{mm} \] and \[ \tan\delta=\frac34 \] \[ \delta=\tan^{-1}\left(\frac34\right) \] \(\textbf{Locating the first positive resultant maximum:}\) At \(t=0\), \[ y(x,0)=5\cos(kx-\delta)\quad\text{mm} \] The first positive maximum at \(x\gt 0\) occurs when \[ kx-\delta=0 \] Hence, \[ x=\frac{\delta}{k} \] The wavelength is \(4.0\,\text{m}\), so \[ k=\frac{2\pi}{4.0}=\frac{\pi}{2}\,\text{rad m}^{-1} \] Therefore, \[ x=\frac{2}{\pi}\tan^{-1}\left(\frac34\right)\,\text{m} \] \(\textbf{Transverse particle velocity at the origin:}\) The frequency is \[ f=2.0\,\text{Hz} \] Thus, \[ \omega=2\pi f=4\pi\,\text{rad s}^{-1} \] Differentiate the original superposition: \[ u_y=-3\omega\cos\phi+4\omega\sin\phi\quad\text{mm s}^{-1} \] At \(x=0,t=0\), the phase is zero. \[ u_y=-3(4\pi) \] \[ u_y=-12\pi\,\text{mm s}^{-1} \] \(\textbf{Physical interpretation:}\) The component amplitudes cannot be added directly because the waves differ in phase by one-quarter cycle. Their resultant has an amplitude determined by the vector-like addition of sine and cosine coefficients. The phase of the resultant determines where its first crest occurs, while the temporal derivative determines the motion of a selected string particle. The superposition still travels rightwards with the original frequency and wavelength because both component waves share those quantities. \(\textbf{Final answer:}\) \[ \boxed{R=5\,\text{mm},\qquad x_{\mathrm{first\ crest}}=\frac2\pi\tan^{-1}\left(\frac34\right)\,\text{m},\qquad u_y(0,0)=-12\pi\,\text{mm s}^{-1}} \]

Explanation: \(\textbf{Maximum resultant amplitude:}\) Let the component amplitudes be \(A_1\) and \(A_2\), with \(A_1\gt A_2\). When the waves are in phase, their displacements reinforce one another at every corresponding phase. The largest resultant amplitude is \[ R_{\max}=A_1+A_2 \] Therefore, \[ A_1+A_2=7.0\,\text{cm} \] \(\textbf{Minimum resultant amplitude:}\) When the phase difference is \(\pi\), the component displacements oppose one another. The minimum resultant amplitude is \[ R_{\min}=|A_1-A_2| \] Since \(A_1\gt A_2\), \[ A_1-A_2=1.0\,\text{cm} \] \(\textbf{Solving for the individual amplitudes:}\) Add the two equations: \[ 2A_1=8.0\,\text{cm} \] \[ A_1=4.0\,\text{cm} \] Subtract to obtain \[ 2A_2=6.0\,\text{cm} \] \[ A_2=3.0\,\text{cm} \] \(\textbf{Resultant amplitude at a quarter-cycle phase difference:}\) For two harmonic waves of the same frequency travelling in the same direction, \[ R^2=A_1^2+A_2^2+2A_1A_2\cos\delta \] When \[ \delta=\frac{\pi}{2} \] the interference term vanishes because \(\cos(\pi/2)=0\). Thus, \[ R^2=4.0^2+3.0^2 \] \[ R^2=25.0\,\text{cm}^2 \] \[ R=5.0\,\text{cm} \] \(\textbf{Physical interpretation:}\) The largest and smallest possible resultant amplitudes reveal the sum and difference of the individual amplitudes. They therefore allow the individual waves to be reconstructed without measuring them separately. At a phase difference of \(\pi/2\), neither complete reinforcement nor maximum cancellation occurs. The resultant amplitude lies between the two limiting values. The minimum amplitude is not zero because the two component amplitudes are unequal. \(\textbf{Final answer:}\) \[ \boxed{A_1=4.0\,\text{cm},\qquad A_2=3.0\,\text{cm},\qquad R_{\delta=\pi/2}=5.0\,\text{cm}} \]

Explanation: \(\textbf{Superposing displacements during complete overlap:}\) The two individual wave displacements are \[ y_1=f(x-vt) \] and \[ y_2=-f(x+vt) \] The total displacement is \[ y=y_1+y_2 \] At \(t=0\), \[ y(x,0)=f(x)-f(x) \] \[ y(x,0)=0 \] Thus, the string is instantaneously straight throughout the overlapping portions. \(\textbf{Velocity of the right-moving contribution:}\) The transverse particle velocity associated with the first pulse is \[ u_1=\frac{\partial y_1}{\partial t} \] Differentiating, \[ u_1=-vf'(x-vt) \] At \(t=0\), \[ u_1=-vf'(x) \] \(\textbf{Velocity of the left-moving contribution:}\) For the second pulse, \[ u_2=\frac{\partial y_2}{\partial t} \] Differentiating, \[ u_2=-vf'(x+vt) \] At \(t=0\), \[ u_2=-vf'(x) \] The two velocity contributions are in the same direction, despite the displacements being opposite. \(\textbf{Resultant particle velocity:}\) By linear superposition, \[ u=u_1+u_2 \] \[ u=-2vf'(x) \] Substituting \(v=5.0\,\text{m s}^{-1}\) and \(f'(x)=+0.10\), \[ u=-2(5.0)(0.10) \] \[ u=-1.0\,\text{m s}^{-1} \] \(\textbf{Mechanical-energy interpretation:}\) The resultant displacement and spatial slope vanish at this instant, but the transverse particle velocity is nonzero. Consequently, the local kinetic energy density \[ u_K=\frac12\mu u^2 \] remains positive wherever the profile slope is nonzero. The pulses have not been destroyed. \(\textbf{Physical interpretation:}\) Destructive displacement superposition does not necessarily produce zero particle motion. At this particular instant, the two pulse shapes cancel geometrically while their induced particle velocities reinforce. The energy that was carried by the approaching disturbances remains in the string's motion. Immediately afterwards, the pulses continue propagating through one another in the stipulated ideal linear medium. This distinguishes temporary displacement cancellation from physical annihilation of wave energy. \(\textbf{Final answer:}\) \[ \boxed{y=0,\qquad u_y=-1.0\,\text{m s}^{-1},\qquad u_K\gt 0} \]

Explanation: \(\textbf{Identifying the leading and trailing edges:}\) The rectangular pulse initially occupies \[ 2.0\lt x\lt 3.0\,\text{m} \] Since it moves rightwards, its leading edge is at \(3.0\,\text{m}\). Its trailing edge is at \(2.0\,\text{m}\). The fixed boundary is at \[ x_b=5.0\,\text{m} \] \(\textbf{First contact with the boundary:}\) The leading edge travels a distance \[ d_{\mathrm{first}}=5.0-3.0 \] \[ d_{\mathrm{first}}=2.0\,\text{m} \] Therefore, \[ t_{\mathrm{first}}=\frac{d_{\mathrm{first}}}{v} \] \[ t_{\mathrm{first}}=\frac{2.0}{2.0} \] \[ t_{\mathrm{first}}=1.0\,\text{s} \] \(\textbf{Completion of reflection:}\) The trailing edge must also reach the boundary. Its initial distance from the boundary is \[ d_{\mathrm{last}}=5.0-2.0 \] \[ d_{\mathrm{last}}=3.0\,\text{m} \] Hence, \[ t_{\mathrm{last}}=\frac{3.0}{2.0} \] \[ t_{\mathrm{last}}=1.5\,\text{s} \] \(\textbf{Position of the reflected leading edge:}\) At \(t=2.0\,\text{s}\), the part reflected first has travelled leftwards for \[ 2.0-1.0=1.0\,\text{s} \] It is therefore located at \[ x_1=5.0-2.0(1.0) \] \[ x_1=3.0\,\text{m} \] \(\textbf{Position of the reflected trailing edge:}\) The last-reflected portion has travelled leftwards for \[ 2.0-1.5=0.5\,\text{s} \] Its position is \[ x_2=5.0-2.0(0.5) \] \[ x_2=4.0\,\text{m} \] The reflected pulse therefore occupies \[ 3.0\lt x\lt 4.0\,\text{m} \] \(\textbf{Sign of the reflected displacement:}\) A transverse displacement pulse reflects with inversion at an ideal fixed end. Thus, the originally upward pulse becomes downward. Its signed height is \[ y_{\mathrm{reflected}}=-2.0\,\text{cm} \] \(\textbf{Physical interpretation:}\) The pulse does not reflect instantaneously as a complete object when its leading edge first touches the boundary. Different parts of the pulse reach the end at different times. The reflected pulse is fully formed only after the trailing edge has reached the boundary. The fixed-end condition requires zero transverse displacement at the endpoint, producing an inverted reflected pulse. After complete reflection, the pulse moves leftwards with its original width and speed in the idealised model. \(\textbf{Final answer:}\) \[ \boxed{t_{\mathrm{first}}=1.0\,\text{s},\qquad t_{\mathrm{last}}=1.5\,\text{s},\qquad 3.0\lt x\lt 4.0\,\text{m},\quad y=-2.0\,\text{cm}} \]

Explanation: Reflection at a string boundary is governed by the mechanical condition imposed on that boundary. At a rigidly fixed end, the string endpoint cannot undergo transverse displacement. The reflected displacement pulse must therefore be inverted relative to the incident pulse so that their displacements cancel at the endpoint. This makes Statements I and II correct for string P. At an ideal freely moving end, the transverse force on the endpoint vanishes. In the small-slope approximation, this condition requires the total transverse spatial slope to be zero at the end. A non-inverted reflected displacement pulse satisfies this free-end condition. Its displacement reinforces that of the incident pulse at the endpoint, while the corresponding spatial slopes cancel, making Statement III correct. The two boundaries thus impose different conditions: zero displacement for a fixed end and zero transverse slope for a free end. In an ideal lossless reflection, the pulse energy is not destroyed or absorbed merely because the pulse reverses its propagation direction. Once the incident and reflected portions separate, the returning pulses can retain their original displacement amplitudes. Statement IV is therefore incorrect. The temporary cancellation or reinforcement at a boundary must be distinguished from the amplitude of the fully reflected travelling pulse.

Explanation: \(\textbf{Propagation directions at the endpoint:}\) The incident displacement is \[ y_i(x,t)=A\cos(kx+\omega t) \] A constant incident phase satisfies \[ kx+\omega t=\text{constant} \] Therefore, \[ \frac{dx}{dt}=-\frac{\omega}{k} \] The wave travels towards negative \(x\), correctly approaching the fixed end of the string occupying \(x\geq0\). The reflected wave travels towards positive \(x\). \(\textbf{Applying the fixed-end displacement condition:}\) The boundary at \(x=0\) cannot move transversely. Therefore, \[ y_i(0,t)+y_r(0,t)=0 \] At the boundary, \[ y_i(0,t)=A\cos(\omega t) \] A compatible reflected-wave equation is \[ y_r(x,t)=-A\cos(kx-\omega t) \] This wave propagates towards positive \(x\). \(\textbf{Verifying the boundary condition:}\) At \(x=0\), \[ y_r(0,t)=-A\cos(-\omega t) \] \[ y_r(0,t)=-A\cos(\omega t) \] Consequently, \[ y(0,t)=A\cos\omega t-A\cos\omega t \] \[ y(0,t)=0 \] \(\textbf{Displacement at the specified event:}\) At \[ x=\frac{\lambda}{4},\qquad t=\frac T4 \] the spatial and temporal phase contributions are \[ kx=\frac{\pi}{2},\qquad \omega t=\frac{\pi}{2} \] Thus, \[ y_i=A\cos\left(\frac{\pi}{2}+\frac{\pi}{2}\right) \] \[ y_i=-A \] For the reflected wave, \[ y_r=-A\cos\left(\frac{\pi}{2}-\frac{\pi}{2}\right) \] \[ y_r=-A \] \(\textbf{Resultant displacement:}\) \[ y=y_i+y_r \] \[ y=-A-A \] \[ y=-2A \] \(\textbf{Physical interpretation:}\) The fixed endpoint always has zero displacement because the incident and reflected waves cancel there. Away from that endpoint, the two contributions can reinforce or oppose one another according to their phases. At the specified position and time, both contributions are negative, so their sum reaches \(-2A\). The reflected wave retains the incident frequency and wavelength but travels in the opposite direction. The inversion of its displacement is required by the rigid boundary. \(\textbf{Final answer:}\) \[ \boxed{y_r=-A\cos(kx-\omega t),\qquad y\left(\frac{\lambda}{4},\frac T4\right)=-2A,\qquad y(0,t)=0} \]

Explanation: \(\textbf{Representing the original pulse profile:}\) Let \(f(\xi)\) denote the triangular displacement profile at \(t=0\), in centimetres. The original profile has its maximum at \(\xi=-3.0\,\text{m}\). For the given one-metre rising and falling sections, \[ f(\xi)=3.0[1-|\xi+3.0|]\,\text{cm} \] when \(-4.0\lt \xi\lt -2.0\,\text{m}\), and zero elsewhere. \(\textbf{Incident displacement at a later time:}\) The pulse travels rightwards at \(v=2.0\,\text{m s}^{-1}\). Therefore, \[ y_i(x,t)=f(x-vt) \] \(\textbf{Reflected displacement at the fixed boundary:}\) A fixed boundary requires the resultant displacement to vanish at \(x=0\). The ideal inverted reflected pulse can be represented as \[ y_r(x,t)=-f(-x-vt) \] Thus, throughout the physical string, \[ y(x,t)=f(x-vt)-f(-x-vt) \] At the endpoint this gives \[ y(0,t)=f(-vt)-f(-vt)=0 \] \(\textbf{Evaluating the incident contribution:}\) At \[ x=-0.25\,\text{m},\qquad t=1.25\,\text{s} \] the translation distance is \[ vt=2.0(1.25)=2.50\,\text{m} \] The incident profile coordinate is \[ \xi_i=x-vt \] \[ \xi_i=-0.25-2.50=-2.75\,\text{m} \] Therefore, \[ y_i=3.0[1-|-2.75+3.0|] \] \[ y_i=3.0(0.75) \] \[ y_i=+2.25\,\text{cm} \] \(\textbf{Evaluating the reflected contribution:}\) The reflected profile coordinate is \[ \xi_r=-x-vt \] \[ \xi_r=+0.25-2.50=-2.25\,\text{m} \] The original-profile value there is \[ f(-2.25)=3.0[1-|-2.25+3.0|] \] \[ f(-2.25)=3.0(0.25) \] \[ f(-2.25)=+0.75\,\text{cm} \] The fixed-end reflection reverses its sign: \[ y_r=-0.75\,\text{cm} \] \(\textbf{Resultant displacement:}\) \[ y=y_i+y_r \] \[ y=2.25-0.75 \] \[ y=+1.50\,\text{cm} \] \(\textbf{Physical interpretation:}\) At the specified time, the pulse has only partly reflected. The displacement near the endpoint is therefore the algebraic sum of the still-arriving incident portion and the inverted reflected portion. The fixed endpoint remains stationary because the two contributions cancel exactly there. At a point away from the endpoint, their magnitudes need not be equal, so partial cancellation leaves a nonzero displacement. The result cannot be obtained by simply moving the original triangle backwards after its first contact with the boundary. \(\textbf{Final answer:}\) \[ \boxed{y(-0.25\,\text{m},1.25\,\text{s})=+1.50\,\text{cm},\qquad y(0,1.25\,\text{s})=0} \]

Explanation: The reflected form of a sound disturbance is determined by the boundary condition, not merely by the fact that the propagation direction reverses. At an ideal rigid closed end, the fluid cannot move through the boundary, so the longitudinal particle velocity is zero there. The incident and reflected particle-velocity contributions must cancel at the boundary. For a pressure wave, this condition gives a reflected excess-pressure pulse of the same sign as the incident pressure pulse. Thus, a compression reflects as a compression at the rigid end, making Statement I correct. At an ideal open end, the pressure is constrained to remain at the surrounding atmospheric value. The total excess pressure must therefore vanish at that boundary. An incident compression requires a reflected rarefaction to satisfy this condition, making Statement II correct. Statement III states the two corresponding boundary conditions and is also correct. Although the pressure reflection behaviour differs at the two ends, reflected disturbances travel back into the original medium in both experiments. Reversing propagation direction alone does not determine whether the pressure perturbation changes sign. Statement IV is therefore incorrect. The distinction between pressure and particle displacement is especially important in organ pipes, because a pressure node and a displacement node are not the same boundary condition.

Explanation: \(\textbf{Wavelength of the coherent waves:}\) Both waves have the same frequency and propagation speed. \[ f=50\,\text{Hz},\qquad v=100\,\text{m s}^{-1} \] Therefore, \[ \lambda=\frac vf \] \[ \lambda=\frac{100}{50} \] \[ \lambda=2.0\,\text{m} \] \(\textbf{Resultant amplitude at P:}\) For two harmonic waves of equal amplitude \(A\), the resultant displacement amplitude is \[ R=2A\left|\cos\frac{\delta}{2}\right| \] At P, the second wave lags by \[ \delta_P=\frac{\pi}{3} \] Using \(A=2.0\,\text{cm}\), \[ R_P=2(2.0)\cos\left(\frac{\pi}{6}\right) \] \[ R_P=2\sqrt3\,\text{cm} \] \(\textbf{Additional phase lag caused by the longer path:}\) At Q, the second wave travels an additional distance \[ \Delta l=\frac13\,\text{m} \] This produces an additional phase lag \[ \Delta\delta=\frac{2\pi\Delta l}{\lambda} \] Substituting, \[ \Delta\delta=\frac{2\pi(1/3)}{2.0} \] \[ \Delta\delta=\frac{\pi}{3} \] \(\textbf{Relative phase at Q:}\) Because the extra path belongs to the already lagging second wave, the lags add. \[ \delta_Q=\delta_P+\Delta\delta \] \[ \delta_Q=\frac{\pi}{3}+\frac{\pi}{3} \] \[ \delta_Q=\frac{2\pi}{3} \] \(\textbf{Resultant amplitude at Q:}\) \[ R_Q=2A\left|\cos\frac{\delta_Q}{2}\right| \] \[ R_Q=4.0\left|\cos\frac{\pi}{3}\right| \] \[ R_Q=2.0\,\text{cm} \] \(\textbf{Comparing the interference conditions:}\) The resultant amplitude at P is \(2\sqrt3\,\text{cm}\), whereas at Q it is \(2.0\,\text{cm}\). Therefore, interference is more destructive at Q. Nevertheless, cancellation is incomplete because the relative phase is not \(\pi\). \(\textbf{Physical interpretation:}\) Path difference changes the relative phase of two coherent waves. Its effect must be combined with any phase difference already present at the first observation point. The direction of the extra phase change matters, which is why the stated initial lag determines the resulting phase change. The two waves retain their individual amplitudes and frequencies, but their resultant amplitude changes as their relative phase changes. \(\textbf{Final answer:}\) \[ \boxed{R_P=2\sqrt3\,\text{cm},\qquad R_Q=2.0\,\text{cm},\qquad R_Q\lt R_P} \]

Explanation: \(\textbf{Propagation direction of the incident wave:}\) The incident phase is \[ kx+\omega t \] A constant phase moves towards decreasing \(x\). Thus, the wave approaches the endpoint at \(x=0\) from the physical string occupying \(x\geq0\). The reflected wave must move towards positive \(x\). \(\textbf{Reflection at the fixed endpoint:}\) At a perfectly fixed endpoint, the total transverse displacement must vanish at every instant. The incident displacement there is \[ y_i(0,t)=A\sin(\omega t) \] A compatible reflected wave is \[ y_{r,\mathrm{fixed}}=-A\sin(\omega t-kx) \] Therefore, at the endpoint, \[ y_{\mathrm{fixed}}(0,t)=A\sin\omega t-A\sin\omega t \] \[ y_{\mathrm{fixed}}(0,t)=0 \] Since the endpoint remains at rest, \[ a_{\mathrm{fixed}}(0,t)=0 \] \(\textbf{Reflection at the free endpoint:}\) At an ideal free end, the total transverse force must vanish. In the small-slope approximation, this requires \[ \left.\frac{\partial y}{\partial x}\right|_{x=0}=0 \] A compatible non-inverted reflected displacement is \[ y_{r,\mathrm{free}}=A\sin(\omega t-kx) \] The spatial slopes of the incident and reflected waves cancel at the boundary. \(\textbf{Free-end displacement:}\) At \(x=0\), \[ y_{\mathrm{free}}(0,t)=A\sin\omega t+A\sin\omega t \] \[ y_{\mathrm{free}}(0,t)=2A\sin\omega t \] At \(t=T/4\), \[ \omega t=\frac{\pi}{2} \] Therefore, \[ y_{\mathrm{free}}(0,T/4)=2A \] \(\textbf{Free-end transverse acceleration:}\) Differentiate the endpoint displacement twice with respect to time. \[ a_{\mathrm{free}}(0,t)=-2A\omega^2\sin\omega t \] Thus, \[ a_{\mathrm{free}}(0,T/4)=-2A\omega^2 \] \(\textbf{Physical interpretation:}\) A fixed boundary prevents transverse motion and therefore maintains zero endpoint displacement and acceleration. A free boundary permits transverse motion but requires cancellation of the spatial slopes associated with the transverse tension force. The two contributions consequently reinforce the endpoint displacement at a free end. At the specified quarter-period instant, the free endpoint is at its positive displacement extreme and has maximum downward restoring acceleration. The difference arises entirely from the boundary conditions, not from a difference in incident-wave amplitude. \(\textbf{Final answer:}\) \[ \begin{aligned} \text{Fixed end:}&\quad y=0,\quad a_y=0\\ \text{Free end:}&\quad y=2A,\quad a_y=-2A\omega^2 \end{aligned} \]

Explanation: \(\textbf{Direct propagation from source to sensor:}\) The source is at \[ x_s=0 \] and the sensor is at \[ x_d=10\,\text{m} \] The direct propagation distance is \[ d_i=10\,\text{m} \] Using sound speed \(v=300\,\text{m s}^{-1}\), \[ t_i=\frac{d_i}{v} \] \[ t_i=\frac{10}{300} \] \[ t_i=\frac1{30}\,\text{s} \] \(\textbf{Path travelled by the reflected pulse:}\) The pulse first travels from the source to the rigid end at \(x=30\,\text{m}\). That path has length \[ d_1=30\,\text{m} \] After reflection, it returns from the rigid end to the sensor at \(x=10\,\text{m}\). The return-path length is \[ d_2=30-10 \] \[ d_2=20\,\text{m} \] \(\textbf{Total reflected-pulse travel time:}\) The total path length is \[ d_r=d_1+d_2 \] \[ d_r=30+20 \] \[ d_r=50\,\text{m} \] Therefore, \[ t_r=\frac{d_r}{v} \] \[ t_r=\frac{50}{300} \] \[ t_r=\frac16\,\text{s} \] \(\textbf{Time interval between detections:}\) \[ \Delta t=t_r-t_i \] \[ \Delta t=\frac16-\frac1{30} \] \[ \Delta t=\frac5{30}-\frac1{30} \] \[ \Delta t=\frac2{15}\,\text{s} \] \(\textbf{Sign of the reflected pressure disturbance:}\) At the rigid closed end, the longitudinal particle velocity must vanish. The incident and reflected particle-velocity contributions therefore cancel at the boundary. Their excess-pressure contributions instead have the same sign. An incident compression consequently reflects as a compression. \(\textbf{Physical interpretation:}\) The sensor receives the incident pulse before the pulse reaches the closed end. The echo follows a longer path consisting of the outward journey to the boundary and the return journey to the sensor. The separation between detections is determined by the additional path travelled by the echo. The reflected pressure pulse retains its positive sign at a rigid end even though its propagation direction reverses. This distinguishes reflection of pressure disturbances from reflection of transverse displacement pulses at fixed string ends. \(\textbf{Final answer:}\) \[ \boxed{t_i=\frac1{30}\,\text{s},\qquad t_r=\frac16\,\text{s},\qquad \Delta t=\frac2{15}\,\text{s},\qquad p'_{\mathrm{reflected}}\gt 0} \]

Explanation: \(\textbf{Arrival of the pulse edges at the free end:}\) The leading edge initially lies at \[ x_{\mathrm{lead},0}=-1.0\,\text{m} \] It reaches the endpoint after \[ t_{\mathrm{first}}=\frac{0-(-1.0)}{2.0} \] \[ t_{\mathrm{first}}=0.50\,\text{s} \] The trailing edge starts at \(-3.0\,\text{m}\), so \[ t_{\mathrm{last}}=\frac{0-(-3.0)}{2.0} \] \[ t_{\mathrm{last}}=1.50\,\text{s} \] \(\textbf{Incident pulse at }t=1.00\,\textbf{s:}\) The incident profile translates by \[ vt=2.0(1.00)=2.0\,\text{m} \] Its mathematical interval becomes \[ -1.0\lt x\lt 1.0 \] Only the portion \(-1.0\lt x\lt 0\) belongs to the physical string. \(\textbf{Reflected portion at the same instant:}\) Reflection begins at \(t=0.50\,\text{s}\). By \(t=1.00\,\text{s}\), the first-reflected portion has travelled leftwards through \[ d=2.0(1.00-0.50) \] \[ d=1.0\,\text{m} \] Thus, the reflected portion occupies \[ -1.0\lt x\lt 0 \] \(\textbf{Applying the free-end condition:}\) At an ideal free end, a transverse displacement pulse reflects without inversion. The overlapping displacements therefore add: \[ y_{\mathrm{resultant}}=+2.0+2.0 \] \[ y_{\mathrm{resultant}}=+4.0\,\text{cm} \] \(\textbf{Position of the fully reflected pulse at }t=1.75\,\textbf{s:}\) Reflection is complete after \(1.50\,\text{s}\). The portion reflected first has travelled leftwards for \[ 1.75-0.50=1.25\,\text{s} \] Its position is \[ x_1=-2.0(1.25)=-2.5\,\text{m} \] The portion reflected last has travelled for \[ 1.75-1.50=0.25\,\text{s} \] Its position is \[ x_2=-2.0(0.25)=-0.5\,\text{m} \] Therefore, the fully reflected pulse occupies \[ -2.5\lt x\lt -0.5 \] Its height returns to the original \(+2.0\,\text{cm}\). \(\textbf{Physical interpretation:}\) The displacement at a free end can be twice the incident pulse displacement during reflection. This temporary reinforcement does not mean that the returning pulse permanently doubles its amplitude. After the two travelling contributions separate, the reflected pulse retains its original height and width. The calculation distinguishes the displacement produced by overlap from the amplitude of one independently travelling pulse. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} t=1.00\,\text{s}:\ -1.0\lt x\lt 0,\quad y=+4.0\,\text{cm}\\ t=1.75\,\text{s}:\ -2.5\lt x\lt -0.5,\quad y=+2.0\,\text{cm} \end{gathered}} \]

Explanation: \(\textbf{Superposing the counter-propagating waves:}\) The total transverse displacement is \[ y=A\sin(kx-\omega t)+A\sin(kx+\omega t) \] Using the sum-of-sines identity, \[ y=2A\sin(kx)\cos(\omega t) \] The opposite travelling directions produce a time-dependent displacement pattern whose spatial factor is fixed. \(\textbf{Wave number and observation phase:}\) \[ k=\frac{2\pi}{\lambda} \] \[ k=\frac{2\pi}{0.80} \] \[ k=\frac{5\pi}{2}\,\text{rad m}^{-1} \] At \(x=0.10\,\text{m}\), \[ kx=\frac{5\pi}{2}(0.10) \] \[ kx=\frac{\pi}{4} \] \(\textbf{Resultant displacement:}\) At \(t=0\), \[ y=2A\sin\left(\frac{\pi}{4}\right)\cos0 \] \[ y=\sqrt2 A \] \[ y=\sqrt2\,\text{cm} \] \(\textbf{Particle velocity at the observation instant:}\) Differentiate the resultant displacement at fixed \(x\): \[ u_y=-2A\omega\sin(kx)\sin(\omega t) \] At \(t=0\), \[ u_y=0 \] Both travelling contributions induce opposite transverse velocities that cancel. \(\textbf{Elastic energy stored in the local string slope:}\) The spatial derivative is \[ \frac{\partial y}{\partial x}=2Ak\cos(kx)\cos(\omega t) \] At the specified event, \[ \frac{\partial y}{\partial x}=2Ak\cos\left(\frac{\pi}{4}\right) \] \[ \frac{\partial y}{\partial x}=\sqrt2 Ak \] Since transverse velocity is zero, the kinetic energy density is zero. The elastic contribution is \[ u_E=\frac12F(\sqrt2 Ak)^2 \] \[ u_E=FA^2k^2 \] Substituting, \[ u_E=32(0.010)^2\left(\frac{5\pi}{2}\right)^2 \] \[ u_E=0.020\pi^2\,\text{J m}^{-1} \] \(\textbf{Physical interpretation:}\) At this instant, the particle is at rest even though its displacement is nonzero. The two oppositely travelling waves cancel each other's particle velocities while reinforcing their displacement and spatial deformation. Mechanical energy therefore remains stored elastically in the string. Zero instantaneous particle velocity does not imply zero total wave energy. This is different from the cancellation of displacement accompanied by nonzero particle velocity, which can occur for other superpositions. \(\textbf{Final answer:}\) \[ \boxed{y=\sqrt2\,\text{cm},\qquad u_y=0,\qquad u_E=0.020\pi^2\,\text{J m}^{-1}} \]

Explanation: \(\textbf{Verifying the direct pulse arrival:}\) The incident pulse travels from the source at \(x=0\) to the sensor at \(x=6.0\,\text{m}\). The direct travel time is \[ t_i=\frac{6.0}{300} \] \[ t_i=0.020\,\text{s} \] This agrees with the observed first detection. \(\textbf{Path taken by the reflected pulse:}\) The pulse travels from the source to the unknown boundary at \(x=L\). It then returns from that boundary to the sensor. The total reflected path length is \[ d_r=L+(L-6.0) \] \[ d_r=2L-6.0 \] \(\textbf{Recovering the boundary location:}\) The reflected pulse reaches the sensor at \(0.100\,\text{s}\). Hence, \[ 0.100=\frac{2L-6.0}{300} \] Multiplying by \(300\), \[ 30.0=2L-6.0 \] \[ 2L=36.0 \] \[ L=18.0\,\text{m} \] \(\textbf{Checking the separate outward and return times:}\) The pulse reaches the end after \[ t_{\mathrm{end}}=\frac{18.0}{300} \] \[ t_{\mathrm{end}}=0.060\,\text{s} \] The return distance to the sensor is \[ 18.0-6.0=12.0\,\text{m} \] The return time is \[ t_{\mathrm{return}}=\frac{12.0}{300} \] \[ t_{\mathrm{return}}=0.040\,\text{s} \] Thus, \[ t_{\mathrm{end}}+t_{\mathrm{return}}=0.100\,\text{s} \] \(\textbf{Identifying the boundary from pressure inversion:}\) The incident pulse has positive excess pressure and represents compression. At an ideal rigid closed end, a pressure pulse reflects without sign reversal. At an ideal open end maintained at atmospheric pressure, the total excess pressure must vanish. The reflected pressure pulse is consequently inverted. Because the echo has negative excess pressure, the boundary is ideally open. \(\textbf{Detection interval:}\) \[ \Delta t=0.100-0.020 \] \[ \Delta t=0.080\,\text{s} \] \(\textbf{Physical interpretation:}\) The echo arrival time determines the total sound path length, not simply the distance between the sensor and the reflecting boundary. The unknown end position is recovered by separating the outgoing and return paths. The pressure sign provides independent evidence about the mechanical boundary condition. Timing and pulse polarity together identify both the distance and type of reflecting end. \(\textbf{Final answer:}\) \[ L=18.0\,\text{m},\qquad \text{ideal open end},\qquad \Delta t=0.080\,\text{s} \]

Explanation: \(\textbf{Superposing the unequal counter-propagating waves:}\) The total displacement is \[ y=3\sin(kx-\omega t)+2\sin(kx+\omega t) \] Expanding the trigonometric functions, \[ y=3(\sin kx\cos\omega t-\cos kx\sin\omega t) +2(\sin kx\cos\omega t+\cos kx\sin\omega t) \] Collecting like terms, \[ y=5\sin(kx)\cos(\omega t)-\cos(kx)\sin(\omega t) \] Here all displacement coefficients are in centimetres. \(\textbf{Local amplitude at an arbitrary position:}\) At fixed \(x\), this is a linear combination of sine and cosine functions of time. Its oscillation amplitude is \[ R(x)=\sqrt{25\sin^2(kx)+\cos^2(kx)}\,\text{cm} \] \(\textbf{Amplitude at the origin:}\) At \(x=0\), \[ \sin(kx)=0,\qquad \cos(kx)=1 \] Therefore, \[ R(0)=1.0\,\text{cm} \] \(\textbf{Amplitude one-quarter wavelength away:}\) The second position is \[ x=0.30\,\text{m}=\frac{\lambda}{4} \] Hence, \[ kx=\frac{\pi}{2} \] Thus, \[ R(0.30)=5.0\,\text{cm} \] \(\textbf{Particle velocity at the origin:}\) At \(x=0\), the resultant displacement becomes \[ y(0,t)=-\sin(\omega t)\,\text{cm} \] The angular frequency is \[ \omega=2\pi f=10\pi\,\text{rad s}^{-1} \] Differentiating, \[ u_y(0,t)=-\omega\cos(\omega t)\,\text{cm s}^{-1} \] At \(t=0\), \[ u_y(0,0)=-10\pi\,\text{cm s}^{-1} \] \(\textbf{Checking for permanent displacement nodes:}\) Rearrange the local amplitude expression: \[ R^2=1+24\sin^2(kx)\quad\text{cm}^2 \] Since \(\sin^2(kx)\geq0\), \[ R(x)\geq1.0\,\text{cm} \] The amplitude cannot vanish at any position. \(\textbf{Physical interpretation:}\) Equal-frequency counter-propagating waves need not produce exact displacement nodes when their amplitudes differ. The resultant still has position-dependent oscillation amplitudes, but its minimum amplitude is the difference between the component amplitudes. At the origin, the displacements partially cancel while the particle continues oscillating. At one-quarter wavelength, the contributions reinforce to produce their maximum possible resultant amplitude. The unequal-amplitude superposition is therefore not a pure standing wave with permanently stationary nodes. \(\textbf{Final answer:}\) \[ \boxed{R(0)=1.0\,\text{cm},\qquad R(0.30)=5.0\,\text{cm},\qquad u_y(0,0)=-10\pi\,\text{cm s}^{-1}} \] There are no permanently stationary displacement nodes.