Oscillations and Waves MCQs with Answers – Part 4 (NEET Physics)

Explanation: For a longitudinal mechanical wave, the displacement gradient indicates how the spacing between neighbouring medium particles differs from its equilibrium value. A negative spatial gradient indicates locally reduced separation, corresponding to compression. A positive gradient indicates increased separation, corresponding to rarefaction. Therefore, P is a compression and Q is a rarefaction. The propagation direction can be inferred by comparing the signs of particle velocity and the spatial gradient. For a wave travelling towards positive \(x\), displacement has the form \(s=f(x-vt)\), giving \(\partial s/\partial t=-v\,\partial s/\partial x\). The two derivatives must therefore have opposite signs wherever they are nonzero. That condition is not satisfied by the recorded observations. For a wave travelling towards negative \(x\), displacement has the form \(s=f(x+vt)\), giving \(\partial s/\partial t=+v\,\partial s/\partial x\). The derivatives then have the same sign, agreeing with both readings. Hence the wave travels towards negative \(x\). This reasoning does not assume that the medium particles travel bodily along with the wave. It uses the local displacement pattern and its temporal change to identify how the disturbance advances.

Explanation: \(\textbf{Propagation speed from the marked compression:}\) The marked compression travels a distance \[ d=1.05\,\text{m} \] The travel time is \[ \Delta t=3.0\times10^{-3}\,\text{s} \] Therefore, \[ v=\frac d{\Delta t} \] \[ v=\frac{1.05}{3.0\times10^{-3}} \] \[ v=350\,\text{m s}^{-1} \] \(\textbf{Wavelength from compression spacing:}\) In a sinusoidal longitudinal wave, successive compression centres represent equivalent phases. Their separation equals one wavelength. Therefore, \[ \lambda=0.70\,\text{m} \] \(\textbf{Finding the frequency:}\) The wave-speed relation is \[ v=f\lambda \] Rearranging, \[ f=\frac v\lambda \] Substituting, \[ f=\frac{350}{0.70} \] \[ f=500\,\text{Hz} \] \(\textbf{Locating the nearest rarefaction:}\) A compression corresponds to a pressure maximum relative to equilibrium. The adjacent rarefaction corresponds to the following pressure minimum. These positions differ in phase by \(\pi\). A full spatial phase change of \(2\pi\) corresponds to one wavelength, so \[ d_{\mathrm{C-R}}=\frac{\lambda}{2} \] \[ d_{\mathrm{C-R}}=\frac{0.70}{2} \] \[ d_{\mathrm{C-R}}=0.35\,\text{m} \] \(\textbf{Physical interpretation:}\) The travel time of one identified compression provides the wave speed. The simultaneous distance between successive compression centres provides wavelength. These measurements are independent and jointly determine frequency. A rarefaction is not another compression and therefore does not occur one full wavelength from its nearest compression. Instead, the two types of extrema alternate at intervals of half a wavelength. \(\textbf{Final answer:}\) \[ \boxed{v=350\,\text{m s}^{-1},\qquad f=500\,\text{Hz},\qquad d_{\mathrm{C-R}}=0.35\,\text{m}} \]

Explanation: \(\textbf{Wave speed controlled by the medium:}\) For linear sound propagation in unchanged air, the propagation speed is determined by the medium's elastic and inertial properties. The question states that these properties and the temperature remain unchanged. Therefore, \[ v'=v \] \(\textbf{Effect on wavelength:}\) The wave-speed relation is \[ v=f\lambda \] After the source frequency doubles, \[ f'=2f \] Since propagation speed remains unchanged, \[ f'\lambda'=f\lambda \] \[ 2f\lambda'=f\lambda \] Therefore, \[ \lambda'=\frac{\lambda}{2} \] \(\textbf{Particle oscillation speed:}\) At a fixed position, an air particle executes approximately sinusoidal longitudinal motion. Its displacement can be written as \[ s=a\cos(\omega t+\phi) \] The maximum particle-speed magnitude is \[ u_{\max}=a\omega \] Using \(\omega=2\pi f\), \[ u_{\max}=2\pi fa \] \(\textbf{Comparing the particle speeds:}\) The particle displacement amplitude \(a\) is specified to remain unchanged. The frequency doubles. Therefore, \[ \frac{u'_{\max}}{u_{\max}} =\frac{2\pi(2f)a}{2\pi fa} \] \[ \frac{u'_{\max}}{u_{\max}}=2 \] \(\textbf{Physical interpretation:}\) The sound pattern travels through the same air at the same propagation speed after the source frequency changes. Higher frequency places successive compressions closer together, reducing wavelength. However, each air particle completes its oscillation more rapidly. With displacement amplitude held fixed, its maximum oscillation speed doubles. The explicit amplitude condition is essential: without it, changing frequency alone would not uniquely determine the change in particle-speed amplitude. \(\textbf{Final answer:}\) \[ \boxed{\frac{v'}v=1,\qquad \frac{\lambda'}{\lambda}=\frac12,\qquad \frac{u'_{\max}}{u_{\max}}=2} \]

Explanation: \(\textbf{Phase corresponding to maximum compression:}\) Write the wave phase as \[ \phi=kx-\omega t \] The excess pressure is \[ p'=P\cos\phi \] Maximum positive pressure occurs when \[ \cos\phi=1 \] We may choose the representative phase \(\phi=0\). \(\textbf{Particle displacement:}\) The longitudinal displacement is \[ s=-a\sin\phi \] At \(\phi=0\), \[ s=0 \] Thus, a compression centre does not coincide with maximum particle displacement in this travelling wave. \(\textbf{Particle velocity:}\) Particle velocity is the time derivative of displacement at fixed position. \[ u=\frac{\partial s}{\partial t} \] Differentiating, \[ u=a\omega\cos\phi \] At maximum positive pressure, \[ u=+a\omega \] The particle therefore moves towards positive \(x\) at its maximum oscillation speed. \(\textbf{Particle acceleration:}\) Differentiate once more: \[ a_p=\frac{\partial u}{\partial t} \] \[ a_p=a\omega^2\sin\phi \] At \(\phi=0\), \[ a_p=0 \] \(\textbf{Physical interpretation:}\) A compression centre represents a region of maximum excess pressure and density. It does not mean that an individual particle has maximum displacement or acceleration. At this phase, the particle passes through its equilibrium position at maximum speed. The particle acceleration is zero because the local pressure gradient is zero at the pressure maximum. The compression pattern continues travelling through the medium even though the identified particle follows its own oscillatory motion.

Explanation: \(\textbf{Wavelength from the spatial profile:}\) Two successive crests are separated by one wavelength. Therefore, \[ \lambda=0.90-0.20 \] \[ \lambda=0.70\,\text{m} \] The crest-to-adjacent-trough distance is \[ 0.55-0.20=0.35\,\text{m} \] This equals \(\lambda/2\), confirming the wavelength. \(\textbf{Period from propagation speed:}\) For a travelling wave, \[ v=\frac{\lambda}{T} \] Therefore, \[ T=\frac{\lambda}{v} \] Substituting, \[ T=\frac{0.70}{3.5} \] \[ T=0.20\,\text{s} \] \(\textbf{Time elapsed for the marked particle:}\) The requested instant is \[ t=0.050\,\text{s} \] Hence, \[ \frac{t}{T}=\frac{0.050}{0.20} \] \[ \frac{t}{T}=\frac14 \] The marked particle has therefore completed one-quarter of its own transverse oscillation. \(\textbf{Vertical motion of the marked particle:}\) At \(t=0\), this particle is at a crest, so its vertical displacement is \(+A\) and its vertical velocity is zero. For a right-travelling sinusoidal wave, the fixed string particle subsequently descends from that crest. Its displacement can be written locally as \[ y(t)=A\cos(\omega t) \] After one-quarter period, \[ y(T/4)=A\cos\left(\frac{\pi}{2}\right) \] \[ y(T/4)=0 \] The particle is moving downwards through equilibrium at that instant. \(\textbf{Physical interpretation:}\) The wave crest itself advances horizontally along the string. The material particle initially located at the crest does not travel horizontally with it. Instead, that particle oscillates vertically while the wave profile propagates. The crest-to-trough distance supplies half a wavelength, whereas the period describes a complete vertical oscillation of one string particle. \(\textbf{Final answer:}\) \[ \boxed{\lambda=0.70\,\text{m},\qquad T=0.20\,\text{s},\qquad y(T/4)=0,\quad v_y(T/4)\lt 0} \]

Explanation: In Observation P, regions of compression and rarefaction travel through air. Air particles oscillate approximately parallel to the direction of propagation, so the disturbance is longitudinal and Statement I is correct. In Observation Q, the string particles move perpendicular to the equilibrium direction along which the disturbance propagates. This is the defining behaviour of a transverse wave on a stretched string, making Statement II correct. A mechanical wave requires both inertia and a restoring mechanism capable of transferring the disturbance through the medium. A longitudinal pressure wave in a fluid can propagate because the fluid resists compression. A bulk transverse shear wave instead requires a restoring force associated with shear deformation. An ideal fluid has no static shear rigidity and therefore does not support the proposed bulk transverse shear mode; Statement III is correct. This does not exclude transverse disturbances on a stretched string, where tension supplies the restoring action. It also does not mean that every physical disturbance involving a fluid must be purely longitudinal, since surface waves have different boundary conditions and motion. Observation S proposes a mechanical wave through a perfect vacuum, which contains no material particles to provide the required mechanical interactions. Mechanical sound therefore cannot propagate in a perfect vacuum, making Statement IV incorrect. The correct combination is Statements I, II and III only.

Explanation: \(\textbf{Wavelength from the spatial pattern:}\) The displacement is zero at \(x=0\), reaches a crest at \(0.30\,\text{m}\), returns to zero at \(0.60\,\text{m}\), and reaches a trough at \(0.90\,\text{m}\). These positions correspond to successive quarter-wavelength intervals. Therefore, \[ \frac{\lambda}{4}=0.30\,\text{m} \] \[ \lambda=1.20\,\text{m} \] \(\textbf{Period from wave speed:}\) \[ v=\frac{\lambda}{T} \] Therefore, \[ T=\frac{\lambda}{v} \] \[ T=\frac{1.20}{3.0} \] \[ T=0.40\,\text{s} \] \(\textbf{Identifying propagation direction:}\) The recorded displacement profile increases with \(x\) near the origin. Thus, its local spatial slope is positive. The particle at the origin is simultaneously moving upwards, so its transverse velocity is positive. For a right-travelling wave, \[ y=f(x-vt) \] Differentiation gives \[ \frac{\partial y}{\partial t} =-v\frac{\partial y}{\partial x} \] A positive slope would therefore require negative particle velocity. That contradicts the observation. For a left-travelling wave, \[ y=f(x+vt) \] the time and spatial derivatives have the same sign. The wave consequently travels towards negative \(x\). \(\textbf{Displacement after the specified interval:}\) At the origin, the particle initially crosses equilibrium while moving upwards. A suitable local displacement function is \[ y(0,t)=A\sin(\omega t) \] The elapsed time is \[ \Delta t=0.10\,\text{s}=\frac T4 \] Therefore, \[ y(0,T/4)=A\sin\left(\frac{\pi}{2}\right) \] \[ y(0,T/4)=+A \] \(\textbf{Physical interpretation:}\) The spatial pattern determines wavelength but does not, by itself, distinguish leftward from rightward propagation. The observed direction of motion of one string particle resolves that ambiguity. After the wave direction is established, the same particle's time evolution follows its local oscillatory phase. The propagation direction and the particle's direction of movement are not interchangeable quantities. \(\textbf{Final answer:}\) \[ \text{Propagation towards negative }x,\qquad T=0.40\,\text{s},\qquad y(0,0.10\,\text{s})=+A \]

Explanation: \(\textbf{Identifying the wave parameters:}\) The displacement equation has the form \[ y=A\sin(kx-\omega t) \] Therefore, \[ A=0.030\,\text{m} \] \[ k=\frac{10\pi}{3}\,\text{rad m}^{-1} \] \[ \omega=10\pi\,\text{rad s}^{-1} \] \(\textbf{Wave-propagation speed:}\) \[ v_{\mathrm{wave}}=\frac{\omega}{k} \] Substituting, \[ v_{\mathrm{wave}}=\frac{10\pi}{10\pi/3} \] \[ v_{\mathrm{wave}}=3.0\,\text{m s}^{-1} \] The form \(kx-\omega t\) indicates motion towards positive \(x\). \(\textbf{Maximum transverse speed of a particle:}\) The particle velocity is obtained by differentiating displacement with respect to time at fixed \(x\). \[ v_y=\frac{\partial y}{\partial t} \] Thus, \[ v_y=-A\omega\cos(kx-\omega t) \] The maximum magnitude is \[ |v_y|_{\max}=A\omega \] Substituting, \[ |v_y|_{\max}=0.030(10\pi) \] \[ |v_y|_{\max}=0.30\pi\,\text{m s}^{-1} \] \(\textbf{State of the selected particle at }t=0\textbf{:}\) At \(x=0.15\,\text{m}\), \[ kx=\frac{10\pi}{3}(0.15) \] \[ kx=\frac{\pi}{2} \] Therefore, \[ y(0.15,0)=0.030\sin\left(\frac{\pi}{2}\right) \] \[ y(0.15,0)=+0.030\,\text{m} \] The particle is initially at a crest. Its instantaneous transverse velocity is zero at that precise instant. \(\textbf{Direction immediately after the crest:}\) As time increases slightly, the phase at fixed \(x\) decreases from \(\pi/2\). The sine function then decreases from its maximum value. Consequently, the particle moves downwards immediately after \(t=0\). \(\textbf{Physical interpretation:}\) Wave speed describes the horizontal propagation of the displacement pattern. Particle speed describes vertical motion of a particular string element. The wave speed is obtained from the ratio of angular frequency to wave number. The particle-speed amplitude is instead obtained by multiplying displacement amplitude by angular frequency. At a crest, instantaneous particle speed is zero even though the crest itself continues propagating along the string. \(\textbf{Final answer:}\) \[ v_{\mathrm{wave}}=3.0\,\text{m s}^{-1},\qquad |v_y|_{\max}=0.30\pi\,\text{m s}^{-1},\qquad \text{initial motion afterwards is downward} \]

Explanation: \(\textbf{Initial position on the pulse profile:}\) At \(t=0\), the pulse maximum is located at \[ x_{\max}=3.0\,\text{m} \] The marked particle lies at \[ x_P=3.50\,\text{m} \] It is therefore on the falling spatial side of the hump, where displacement decreases as \(x\) increases. Hence, the spatial slope at the marked position is negative. \(\textbf{Relating pulse propagation to particle motion:}\) A pulse travelling towards positive \(x\) has the mathematical form \[ y(x,t)=f(x-vt) \] Differentiating at fixed \(x\), \[ \frac{\partial y}{\partial t} =-v\frac{\partial y}{\partial x} \] Since \(v\gt 0\), the transverse particle velocity has the opposite sign to the spatial slope. The slope is initially negative. Therefore, the marked particle initially moves upwards. \(\textbf{Time for the maximum to reach the particle:}\) The pulse maximum must travel from \(3.0\,\text{m}\) to \(3.50\,\text{m}\). The required distance is \[ \Delta x=3.50-3.0 \] \[ \Delta x=0.50\,\text{m} \] Using the propagation speed, \[ \Delta t=\frac{\Delta x}{v} \] \[ \Delta t=\frac{0.50}{2.0} \] \[ \Delta t=0.25\,\text{s} \] \(\textbf{Motion as the maximum passes:}\) The marked particle continues rising until the pulse maximum reaches its position. At that instant, the spatial slope at the smooth maximum is zero. Its transverse velocity is consequently zero. Immediately afterwards, the particle is on the rising spatial side of the translated pulse profile. That side has positive spatial slope. For rightward propagation, the particle velocity is then negative, so the particle moves downwards. \(\textbf{Physical interpretation:}\) The pulse maximum travels horizontally along the string, but the marked particle remains at its fixed horizontal equilibrium coordinate. Its vertical displacement first increases, reaches a maximum and then decreases. The particle does not travel with the hump. The pulse's propagation speed determines the arrival time, whereas the local profile slope determines the instantaneous direction of transverse particle motion. \(\textbf{Final answer:}\) \[ \text{Initially upward},\qquad t_{\max}=0.25\,\text{s},\qquad \text{afterwards downward} \]

Explanation: \(\textbf{Interpreting the measured graph:}\) The speed of a transverse wave along an ideal stretched string is \[ v=\sqrt{\frac{F}{\mu}} \] where \(F\) is tension and \(\mu\) is linear mass density. Squaring, \[ v^2=\frac{F}{\mu} \] Thus, the slope of the \(v^2\)-versus-\(F\) graph is \(1/\mu\), not \(\mu\). \(\textbf{Calculating the graph slope:}\) Using the two given measurements, \[ S=\frac{400-100}{160-40} \] \[ S=\frac{300}{120} \] \[ S=2.5\,\text{m}^2\text{s}^{-2}\text{N}^{-1} \] Therefore, \[ \frac1\mu=2.5 \] \[ \mu=0.40\,\text{kg m}^{-1} \] \(\textbf{Mass of the specified section:}\) Linear mass density is mass per unit length. \[ \mu=\frac{m}{l} \] Therefore, \[ m=\mu l \] For \(l=0.50\,\text{m}\), \[ m=0.40(0.50) \] \[ m=0.20\,\text{kg} \] \(\textbf{Wave speed at the specified tension:}\) At \(F=40\,\text{N}\), \[ v=\sqrt{\frac{40}{0.40}} \] \[ v=10\,\text{m s}^{-1} \] This agrees with the plotted value \(v^2=100\,\text{m}^2\text{s}^{-2}\). \(\textbf{Wavelength under steady driving:}\) The source frequency is \[ f=20\,\text{Hz} \] Using \[ v=f\lambda \] we obtain \[ \lambda=\frac vf \] \[ \lambda=\frac{10}{20} \] \[ \lambda=0.50\,\text{m} \] \(\textbf{Physical interpretation:}\) Increasing tension increases wave speed, but the relation is a square-root dependence. Squaring the measured speed produces a straight-line relation whose slope directly reveals the reciprocal linear density. The mass of any uniform section can then be calculated from its length. Finally, the source frequency and medium-dependent speed jointly determine wavelength. The graph alone does not determine the source frequency; that independent measurement is essential. \(\textbf{Final answer:}\) \[ \boxed{\mu=0.40\,\text{kg m}^{-1},\qquad m=0.20\,\text{kg},\qquad \lambda=0.50\,\text{m}} \]

Explanation: \(\textbf{Propagation speed in string P:}\) For a stretched string, \[ v=\sqrt{\frac{F}{\mu}} \] In string P, \[ F=100\,\text{N},\qquad \mu_P=0.010\,\text{kg m}^{-1} \] Therefore, \[ v_P=\sqrt{\frac{100}{0.010}} \] \[ v_P=100\,\text{m s}^{-1} \] \(\textbf{Propagation speed in string Q:}\) The tension remains \(100\,\text{N}\), but the linear mass density is greater. \[ v_Q=\sqrt{\frac{100}{0.040}} \] \[ v_Q=50\,\text{m s}^{-1} \] The transmitted wave consequently travels more slowly in Q. \(\textbf{Frequency at the junction:}\) The frequency of the transmitted steady wave is fixed by the source. Therefore, \[ f_Q=f_P=50\,\text{Hz} \] The frequency does not halve merely because wave speed halves. \(\textbf{Wavelength of the transmitted wave:}\) Using \[ v_Q=f_Q\lambda_Q \] we obtain \[ \lambda_Q=\frac{50}{50} \] \[ \lambda_Q=1.0\,\text{m} \] \(\textbf{Time spent travelling through P:}\) \[ t_P=\frac{L_P}{v_P} \] \[ t_P=\frac{2.0}{100} \] \[ t_P=0.020\,\text{s} \] \(\textbf{Time spent travelling through Q:}\) \[ t_Q=\frac{L_Q}{v_Q} \] \[ t_Q=\frac{3.0}{50} \] \[ t_Q=0.060\,\text{s} \] \(\textbf{Total travel time:}\) The wave must traverse the two sections successively. \[ t_{\mathrm{total}}=t_P+t_Q \] \[ t_{\mathrm{total}}=0.020+0.060 \] \[ t_{\mathrm{total}}=0.080\,\text{s} \] \(\textbf{Physical interpretation:}\) The greater linear density of string Q lowers the propagation speed under the same tension. Since the driving frequency remains unchanged at the junction, the transmitted wavelength decreases in the same proportion as speed. The total travel time must be found by adding the times spent in the individual media. Using the average of the two wave speeds over the combined length would not generally give the correct travel time. \(\textbf{Final answer:}\) \[ \boxed{\lambda_Q=1.0\,\text{m},\qquad t_{\mathrm{total}}=0.080\,\text{s}} \]

Explanation: \(\textbf{Original linear mass density:}\) The string initially has mass \(M\) and length \(L\). Therefore, \[ \mu=\frac ML \] Its initial wave speed is \[ v=\sqrt{\frac F\mu} \] \(\textbf{New linear mass density after stretching:}\) The total mass is unchanged, but the length becomes \[ L'=\frac{3L}{2} \] Therefore, \[ \mu'=\frac M{L'} \] \[ \mu'=\frac{M}{3L/2} \] \[ \mu'=\frac{2\mu}{3} \] \(\textbf{New tension and wave speed:}\) The final tension is \[ F'=\frac{9F}{4} \] The final wave speed is \[ v'=\sqrt{\frac{F'}{\mu'}} \] Substituting, \[ v'=\sqrt{\frac{9F/4}{2\mu/3}} \] \[ v'=\sqrt{\frac{27F}{8\mu}} \] Thus, \[ \frac{v'}v=\sqrt{\frac{27}{8}} \] \[ \frac{v'}v=\frac{3\sqrt6}{4} \] \(\textbf{Comparing pulse travel times:}\) For a pulse travelling along the full length, \[ t=\frac Lv \] In the stretched state, \[ t'=\frac{L'}{v'} \] Therefore, \[ \frac{t'}t=\frac{L'}L\frac v{v'} \] Substituting, \[ \frac{t'}t=\frac32\frac{4}{3\sqrt6} \] \[ \frac{t'}t=\frac2{\sqrt6} \] \[ \frac{t'}t=\sqrt{\frac23} \] \(\textbf{Physical interpretation:}\) Stretching changes not only the length the pulse must traverse but also the mass per unit length. The increased tension and decreased linear mass density both increase wave speed. Although the stretched string is longer, its pulse travel time becomes shorter because the speed increase more than compensates for the length increase. Treating linear mass density as constant would neglect an explicitly stated physical change. \(\textbf{Final answer:}\) \[ \boxed{\frac{v'}v=\frac{3\sqrt6}{4},\qquad \frac{t'}t=\sqrt{\frac23}} \]

Explanation: \(\textbf{Wave speed in a fluid:}\) The speed of a longitudinal mechanical wave is \[ v=\sqrt{\frac B\rho} \] where \(B\) is the appropriate bulk modulus and \(\rho\) is mass density. \(\textbf{Comparing medium properties:}\) The second fluid has \[ B_2=\frac{9B_1}{4} \] and \[ \rho_2=4\rho_1 \] Therefore, \[ \frac{v_2}{v_1} =\sqrt{\frac{B_2/\rho_2}{B_1/\rho_1}} \] Substituting, \[ \frac{v_2}{v_1} =\sqrt{\frac{9/4}{4}} \] \[ \frac{v_2}{v_1}=\frac34 \] \(\textbf{Frequency across the interface:}\) The transmitted wave retains the frequency imposed by the source. Thus, \[ f_2=f_1 \] Since \[ v=f\lambda \] the wavelength ratio equals the speed ratio. \[ \frac{\lambda_2}{\lambda_1} =\frac{v_2}{v_1} \] \[ \frac{\lambda_2}{\lambda_1}=\frac34 \] \(\textbf{Comparing the specified travel times:}\) In the first fluid, \[ t_1=\frac d{v_1} \] The distance in the second fluid is \(2d\). Therefore, \[ t_2=\frac{2d}{v_2} \] Taking the ratio, \[ \frac{t_2}{t_1} =\frac{2d/v_2}{d/v_1} \] \[ \frac{t_2}{t_1} =2\frac{v_1}{v_2} \] \[ \frac{t_2}{t_1}=2\left(\frac43\right) \] \[ \frac{t_2}{t_1}=\frac83 \] \(\textbf{Physical interpretation:}\) The second fluid is elastically stiffer but also substantially denser. The density increase dominates in the ratio \(B/\rho\), so the wave travels more slowly. The frequency remains unchanged during transmission, while the wavelength decreases. The longer travel path in the slower medium further increases the travel time. Neither density nor bulk modulus alone is sufficient to predict wave speed. \(\textbf{Final answer:}\) \[ \boxed{\frac{\lambda_2}{\lambda_1}=\frac34,\qquad \frac{t_2}{t_1}=\frac83} \]

Explanation: \(\textbf{Temperature dependence of sound speed:}\) For adiabatic sound propagation in an ideal gas, \[ v=\sqrt{\frac{\gamma RT}{M}} \] Here the gas composition, molar mass \(M\) and ratio of specific heats \(\gamma\) remain unchanged. Therefore, \[ v\propto\sqrt T \] \(\textbf{Sound speed after heating:}\) The initial and final temperatures are \[ T_1=300\,\text{K},\qquad T_2=432\,\text{K} \] Therefore, \[ \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}} \] Substituting, \[ \frac{v_2}{300}=\sqrt{\frac{432}{300}} \] \[ \frac{v_2}{300}=\frac65 \] \[ v_2=360\,\text{m s}^{-1} \] \(\textbf{Wavelength after heating:}\) The loudspeaker frequency remains \[ f=300\,\text{Hz} \] Using \[ v_2=f\lambda_2 \] we obtain \[ \lambda_2=\frac{360}{300} \] \[ \lambda_2=1.20\,\text{m} \] \(\textbf{Original travel time:}\) The sound path length is \[ d=1.8\,\text{m} \] Initially, \[ t_1=\frac d{v_1} \] \[ t_1=\frac{1.8}{300} \] \[ t_1=0.0060\,\text{s} \] \(\textbf{Travel time after heating:}\) \[ t_2=\frac d{v_2} \] \[ t_2=\frac{1.8}{360} \] \[ t_2=0.0050\,\text{s} \] \(\textbf{Reduction in travel time:}\) \[ \Delta t=t_1-t_2 \] \[ \Delta t=0.0060-0.0050 \] \[ \Delta t=0.0010\,\text{s}=1.0\,\text{ms} \] \(\textbf{Physical interpretation:}\) Heating increases the speed of sound in the specified ideal gas because the characteristic elastic response per unit inertia changes with absolute temperature. The fixed source frequency does not change during this process. Consequently, the wavelength increases with the propagation speed. The shorter travel time provides a second experimentally measurable consequence of the heating. Absolute temperature in kelvin must be used in the square-root relation. \(\textbf{Final answer:}\) \[ \boxed{\lambda_2=1.20\,\text{m},\qquad \Delta t=1.0\,\text{ms}} \]

Explanation: The speed of a small-amplitude sound wave depends on the elastic response and density of the medium. For adiabatic disturbances in an ideal gas, the relevant bulk modulus is \(\gamma P\). Consequently, the speed is \(v=\sqrt{\gamma P/\rho}\). During the stated isothermal compression, the absolute temperature and gas composition remain unchanged. Doubling pressure also doubles density, leaving the ratio \(P/\rho\) unchanged. The ratio of specific heats is separately specified to remain constant. Thus, sound speed remains unchanged despite both pressure and density increasing. The same source produces the same frequency after compression, so the relation \(v=f\lambda\) also gives an unchanged wavelength. The fact that the gas is compressed slowly does not mean that the much faster sound disturbances themselves are isothermal. The appropriate adiabatic response is used for the sound wave, whereas the equilibrium gas states are related by an isothermal process. Both the Assertion and Reason are therefore true, and the Reason correctly accounts for the unchanged speed. Together with the fixed source frequency, it also explains the unchanged wavelength.

Explanation: \(\textbf{Elastic and inertial properties of the rods:}\) The longitudinal wave speed in a slender metallic rod is \[ v=\sqrt{\frac Y\rho} \] The material properties satisfy \[ Y_Q=9Y_P \] \[ \rho_Q=4\rho_P \] \(\textbf{Comparing propagation speeds:}\) \[ \frac{v_Q}{v_P} =\sqrt{\frac{Y_Q/\rho_Q}{Y_P/\rho_P}} \] Substituting, \[ \frac{v_Q}{v_P} =\sqrt{\frac94} \] \[ \frac{v_Q}{v_P}=\frac32 \] Thus, the pulse travels faster in Q. \(\textbf{Comparing rod lengths:}\) The specified lengths obey \[ L_Q=2L_P \] A pulse needs time \[ t=\frac Lv \] to reach the far end of a uniform rod. \(\textbf{Comparing end-to-end travel times:}\) \[ \frac{t_Q}{t_P} =\frac{L_Q/v_Q}{L_P/v_P} \] \[ \frac{t_Q}{t_P} =\frac{L_Q}{L_P}\frac{v_P}{v_Q} \] Substituting, \[ \frac{t_Q}{t_P}=2\left(\frac23\right) \] \[ \frac{t_Q}{t_P}=\frac43 \] \(\textbf{Arrival-order inference:}\) Although the pulse speed in Q is greater, the path through Q is twice as long. Its travel time is consequently greater. Since \[ t_Q\gt t_P \] the pulse in rod P arrives first. \(\textbf{Physical interpretation:}\) The speed of a longitudinal mechanical pulse depends on the combined effects of elastic stiffness and mass density. A material with greater Young's modulus need not have a greater wave speed if its density also changes sufficiently. Moreover, the faster wave need not arrive first when the distances travelled differ. The comparison requires separate examination of medium-dependent speed and travel distance. \(\textbf{Final answer:}\) \[ \boxed{\frac{v_Q}{v_P}=\frac32,\qquad \frac{t_Q}{t_P}=\frac43,\qquad t_P\lt t_Q} \]

Explanation: \(\textbf{Original pulse speed from sensor timing:}\) The distance between sensors is \[ d=3.0\,\text{m} \] The initial travel time is \[ t_1=0.060\,\text{s} \] Therefore, \[ v_1=\frac d{t_1} \] \[ v_1=\frac{3.0}{0.060} \] \[ v_1=50\,\text{m s}^{-1} \] \(\textbf{Determining linear mass density:}\) For transverse waves on a stretched string, \[ v_1^2=\frac{F_1}{\mu} \] Therefore, \[ \mu=\frac{F_1}{v_1^2} \] With \(F_1=36\,\text{N}\), \[ \mu=\frac{36}{50^2} \] \[ \mu=0.0144\,\text{kg m}^{-1} \] \(\textbf{New pulse speed:}\) The second travel time is \[ t_2=0.040\,\text{s} \] Therefore, \[ v_2=\frac{3.0}{0.040} \] \[ v_2=75\,\text{m s}^{-1} \] \(\textbf{Tension inferred from the speed change:}\) Since the linear mass density remains constant, \[ F_2=\mu v_2^2 \] Substituting, \[ F_2=0.0144(75)^2 \] \[ F_2=0.0144(5625) \] \[ F_2=81\,\text{N} \] \(\textbf{Wavelength after changing tension:}\) The driven-wave frequency is \[ f=100\,\text{Hz} \] At the new tension, \[ v_2=f\lambda_2 \] Therefore, \[ \lambda_2=\frac{v_2}{f} \] \[ \lambda_2=\frac{75}{100} \] \[ \lambda_2=0.75\,\text{m} \] \(\textbf{Consistency check:}\) The speed increases by a factor of \[ \frac{v_2}{v_1}=\frac{75}{50}=\frac32 \] At constant linear density, tension must increase by the square of this factor: \[ \frac{F_2}{F_1}=\left(\frac32\right)^2=\frac94 \] \[ F_2=36\left(\frac94\right)=81\,\text{N} \] The two methods agree. \(\textbf{Physical interpretation:}\) Pulse timing measures the propagation speed without requiring a sinusoidal source. Combining that speed with known tension determines the string's linear mass density. A second timing measurement then determines an unknown tension independently of the source frequency. Only after the source is introduced is wavelength calculated from speed and frequency. \(\textbf{Final answer:}\) \[ \boxed{\mu=0.0144\,\text{kg m}^{-1},\qquad F_2=81\,\text{N},\qquad \lambda_2=0.75\,\text{m}} \]

Explanation: \(\textbf{Propagation of the frequency-change boundary:}\) The source changes its driving frequency at \(t=0\). This change cannot be communicated instantaneously to distant parts of the string. The boundary between the older and newer wave trains advances at \[ v=40\,\text{m s}^{-1} \] After time \(t\), its position is \[ x_b=vt \] \(\textbf{Position of the boundary after 15 ms:}\) \[ t=15\times10^{-3}\,\text{s} \] Therefore, \[ x_b=40(15\times10^{-3}) \] \[ x_b=0.60\,\text{m} \] \(\textbf{Identifying which frequency reaches each position:}\) The position \(x=0.40\,\text{m}\) lies behind the advancing boundary, nearer the source. It therefore receives the newer \(200\,\text{Hz}\) oscillations. The position \(x=0.80\,\text{m}\) lies ahead of the boundary. It still carries the older \(100\,\text{Hz}\) wave train. \(\textbf{Wavelength of the newer wave train:}\) Using \[ v=f\lambda \] the newer wavelength is \[ \lambda_{\mathrm{new}}=\frac{40}{200} \] \[ \lambda_{\mathrm{new}}=0.20\,\text{m} \] \(\textbf{Wavelength of the older wave train:}\) Similarly, \[ \lambda_{\mathrm{old}}=\frac{40}{100} \] \[ \lambda_{\mathrm{old}}=0.40\,\text{m} \] \(\textbf{Local wave conditions:}\) Thus, near \(x=0.40\,\text{m}\), \[ (f,\lambda)=(200\,\text{Hz},0.20\,\text{m}) \] Near \(x=0.80\,\text{m}\), \[ (f,\lambda)=(100\,\text{Hz},0.40\,\text{m}) \] \(\textbf{Physical interpretation:}\) Changing the source frequency does not instantly alter the oscillation of every string particle. The new driving pattern reaches progressively more distant positions as the disturbance propagates. During the transition, different regions can temporarily contain wave trains produced at different source frequencies. The medium-dependent wave speed remains the same, so wavelength adjusts inversely with frequency. This question distinguishes the source's present state from the state of a distant medium particle that has not yet received the change. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} x=0.40\,\text{m}:&\quad f=200\,\text{Hz},\ \lambda=0.20\,\text{m}\\ x=0.80\,\text{m}:&\quad f=100\,\text{Hz},\ \lambda=0.40\,\text{m} \end{aligned}} \]

Explanation: \(\textbf{Spatial phase difference:}\) For a right-travelling sinusoidal wave, a suitable phase is \[ \phi=kx-\omega t+\phi_0 \] The spatial separation is \[ \Delta x=0.60\,\text{m} \] The wave number is \[ k=\frac{2\pi}{\lambda} \] Thus, the spatial contribution to the phase difference is \[ \Delta\phi_x=\frac{2\pi}{0.80}(0.60) \] \[ \Delta\phi_x=\frac{3\pi}{2} \] \(\textbf{Temporal phase difference:}\) The observation time increases by \[ \Delta t=0.050\,\text{s} \] The angular frequency is \[ \omega=\frac{2\pi}{T} \] For rightward propagation, elapsed time contributes a negative phase change: \[ \Delta\phi_t=-\omega\Delta t \] \[ \Delta\phi_t=-\frac{2\pi}{0.20}(0.050) \] \[ \Delta\phi_t=-\frac{\pi}{2} \] \(\textbf{Net phase change:}\) \[ \Delta\phi=\Delta\phi_x+\Delta\phi_t \] \[ \Delta\phi=\frac{3\pi}{2}-\frac{\pi}{2} \] \[ \Delta\phi=\pi \] \(\textbf{Displacement at the second observation:}\) Two harmonic states separated in phase by \(\pi\) have opposite displacements. Therefore, \[ y_Q=-y_P \] Since \[ y_P=+\frac A2 \] we obtain \[ y_Q=-\frac A2 \] \(\textbf{Direction of transverse particle motion:}\) For equal wave amplitude and angular frequency, a phase difference of \(\pi\) also reverses the sign of transverse particle velocity. The particle at P moves upwards. Therefore, the particle at Q moves downwards at its specified observation instant. \(\textbf{Physical interpretation:}\) The two observations differ in both position and time. Their phase relationship cannot be determined using spatial separation alone or elapsed time alone. For a wave travelling towards positive \(x\), increasing position advances the phase while increasing time decreases it. Combining both changes gives a net phase reversal. The result demonstrates how a spatially displaced observation and a delayed observation must be treated together. \(\textbf{Final answer:}\) \[ \boxed{y_Q=-\frac A2,\qquad v_{y,Q}\lt 0} \]

Explanation: \(\textbf{Relation between local slope and particle velocity:}\) For a wave travelling towards positive \(x\), \[ y(x,t)=f(x-vt) \] Differentiating, \[ \frac{\partial y}{\partial x}=f'(x-vt) \] and \[ \frac{\partial y}{\partial t}=-vf'(x-vt) \] Therefore, \[ \frac{\partial y}{\partial t} =-v\frac{\partial y}{\partial x} \] \(\textbf{Checking propagation direction:}\) At detector P, the spatial gradient is negative while the transverse particle velocity is positive. These signs are opposite, consistent with propagation towards positive \(x\). At Q, the spatial gradient is positive while particle velocity is negative. This also supports propagation towards positive \(x\). \(\textbf{Calculating speed from detector P:}\) Rearranging, \[ v=-\frac{\partial y/\partial t}{\partial y/\partial x} \] Substituting P's measurements, \[ v=-\frac{+0.60}{-0.020} \] \[ v=30\,\text{m s}^{-1} \] \(\textbf{Independent consistency check using Q:}\) \[ v=-\frac{-0.45}{+0.015} \] \[ v=30\,\text{m s}^{-1} \] Both detectors therefore describe a single wave with the same propagation speed and direction. \(\textbf{Tracking the marked crest:}\) A fixed feature of a progressive wave advances with the wave speed. Initially, \[ x_0=0.75\,\text{m} \] After \[ \Delta t=0.080\,\text{s} \] its displacement is \[ \Delta x=v\Delta t \] \[ \Delta x=30(0.080) \] \[ \Delta x=2.40\,\text{m} \] Since propagation is rightward, \[ x_{\mathrm{crest}}=x_0+\Delta x \] \[ x_{\mathrm{crest}}=0.75+2.40 \] \[ x_{\mathrm{crest}}=3.15\,\text{m} \] \(\textbf{Physical interpretation:}\) A string particle's transverse velocity is not the propagation velocity of the wave. However, for a progressive wave of fixed shape, the transverse velocity and local spatial gradient are related through the propagation speed. The two independent measurements provide a useful compatibility test. Once that speed is established, the crest can be tracked without knowing the wave amplitude or frequency. \(\textbf{Final answer:}\) \[ \text{Rightward propagation},\qquad v=30\,\text{m s}^{-1},\qquad x_{\mathrm{crest}}=3.15\,\text{m} \]