Oscillations and Waves MCQs with Answers – Part 6 (NEET Physics)

Explanation: The direction of propagation is important even when two wave superpositions can produce a zero-displacement configuration. In arrangement P, the two waves travel in opposite directions and establish a stationary spatial pattern. Some positions are permanent displacement nodes, while other particles oscillate with position-dependent amplitudes. Statement I is correct. In arrangement Q, the two waves have identical travelling-wave phase dependences except for a phase difference of \(\pi\). Their displacements cancel at every point and time. Each individual component can nevertheless be assigned a nonzero travelling-wave power before superposition, so Statement II is correct. Arrangement P is not physically equivalent to Q because the standing-wave pattern generally contains moving particles and a nonzero resultant displacement away from nodes. It only becomes straight at selected instants. Arrangement Q remains straight at every instant in the stipulated ideal model. Statement III is therefore incorrect. In arrangement P, the equal-amplitude waves transfer equal average powers in opposite directions, so their net average energy flow vanishes. Statement IV is correct. Persistent cancellation in Q requires a constant relative phase difference and equal amplitudes, which is naturally maintained when the waves have the same frequency. Different frequencies would cause their relative phase to change with time. Statement V is incorrect. The comparison distinguishes stationary-wave formation from complete cancellation of co-propagating coherent waves.

Explanation: \(\textbf{Allowed frequencies of the open-open pipe:}\) For a pipe open at both ends, \[ f_n=\frac{nv}{2L} \] The fundamental frequency is \[ f_1=\frac{240}{2(1.00)} \] \[ f_1=120\,\text{Hz} \] Hence, \[ f_n=120n\,\text{Hz} \] \(\textbf{Pressure nodes at the open ends:}\) Both ideal openings are excess-pressure nodes. For the \(n\)th normal mode, a suitable pressure-amplitude factor is \[ P_n(x)\propto\sin\left(\frac{n\pi x}{L}\right) \] A microphone at a pressure node records no pressure oscillation from that mode. \(\textbf{Finding which modes are invisible at the microphone:}\) At \(x=L/3\), \[ P_n(L/3)\propto\sin\left(\frac{n\pi}{3}\right) \] This factor vanishes when \[ \frac{n\pi}{3}=j\pi \] Therefore, \[ n=3j \] The lowest positive invisible mode is the third harmonic. \(\textbf{Its resonance frequency:}\) For \(n=3\), \[ f_3=3f_1 \] \[ f_3=3(120) \] \[ f_3=360\,\text{Hz} \] \(\textbf{Condition recorded by the displacement sensor:}\) Pressure nodes correspond to longitudinal-displacement antinodes in an ideal stationary sound wave. For the third harmonic, a suitable displacement spatial factor is \[ S_3(x)\propto\cos\left(\frac{3\pi x}{L}\right) \] At \(x=L/3\), \[ S_3(L/3)\propto\cos\pi \] \[ \cos\pi=-1 \] Its magnitude is maximum. Thus, the displacement sensor detects strong motion even though the pressure microphone is silent. \(\textbf{Counting the interior pressure nodes:}\) For the third harmonic, pressure nodes occur at \[ x=0,\quad\frac L3,\quad\frac{2L}{3},\quad L \] The endpoints are excluded from the requested count. Therefore, there are two interior pressure nodes. \(\textbf{Physical interpretation:}\) The microphone's failure to detect a resonance does not mean that the pipe cannot support that frequency. The sensor may simply occupy a node of the measured physical quantity. Because pressure and longitudinal displacement have complementary node-antinode patterns, replacing the microphone with a displacement sensor can reveal the same mode strongly. The normal-mode geometry therefore determines which harmonics are visible to a sensor at a particular position. \(\textbf{Final answer:}\) \[ f=360\,\text{Hz},\qquad \text{displacement sensor detects the mode},\qquad N_{\mathrm{interior,\ pressure}}=2 \]

Explanation: \(\textbf{Local oscillation frequencies of the two modes:}\) For a fixed-fixed string, \[ f_n=nf_1 \] With \(f_1=20\,\text{Hz}\), \[ f_2=40\,\text{Hz},\qquad f_5=100\,\text{Hz} \] The corresponding angular frequencies are \[ \omega_2=80\pi,\qquad\omega_5=200\pi\quad\text{rad s}^{-1} \] \(\textbf{Local superposition at the first sensor:}\) The two mode contributions have equal positive amplitudes and begin at positive extremes. Therefore, their local displacements are \[ y_2=a\cos(80\pi t) \] \[ y_5=a\cos(200\pi t) \] Their sum is \[ y=a[\cos(80\pi t)+\cos(200\pi t)] \] \(\textbf{Factorising the local displacement:}\) Using the cosine sum identity, \[ y=2a\cos(140\pi t)\cos(60\pi t) \] Cancellation occurs whenever either cosine factor is zero. \(\textbf{Finding the earliest positive cancellation:}\) The first zero of \(\cos(140\pi t)\) occurs when \[ 140\pi t=\frac{\pi}{2} \] \[ t=\frac1{280}\,\text{s} \] The first zero of \(\cos(60\pi t)\) is at \[ t=\frac1{120}\,\text{s} \] Since \(1/280\lt 1/120\), the earliest cancellation is \[ t_{\mathrm{first}}=\frac1{280}\,\text{s} \] \(\textbf{Individual displacements at that instant:}\) For the second harmonic, \[ 80\pi t=\frac{2\pi}{7} \] Therefore, \[ y_2=a\cos\left(\frac{2\pi}{7}\right) \] For the fifth harmonic, \[ 200\pi t=\frac{5\pi}{7} \] Hence, \[ y_5=a\cos\left(\frac{5\pi}{7}\right) \] \[ y_5=-a\cos\left(\frac{2\pi}{7}\right) \] The contributions cancel while each remains nonzero. \(\textbf{Checking the midpoint's spatial factors:}\) For a string fixed at both ends, \[ S_n(x)\propto\sin\left(\frac{n\pi x}{L}\right) \] At \(x=L/2\), \[ S_2(L/2)\propto\sin\pi=0 \] \[ S_5(L/2)\propto\sin\frac{5\pi}{2}=1 \] Thus, the midpoint sensor detects the fifth mode but not the second. \(\textbf{Physical interpretation:}\) At the first sensor, two different normal-mode frequencies combine to produce occasional zero-displacement instants. At the earliest such instant, the two modes have equal and opposite nonzero displacements rather than each passing through equilibrium. The midpoint acts as a spatial filter because it is a node of one mode but not of the other. The earliest local cancellation occurs when the two mode contributions have equal magnitudes and opposite signs, although neither contribution is individually zero. The midpoint is a permanent node of the second harmonic but an antinode of the fifth harmonic. Thus, cancellation at the first sensor does not imply cancellation at the midpoint. \(\textbf{Final answer:}\) \[ \boxed{t_{\mathrm{first}}=\frac1{280}\,\text{s},\qquad y_2=+a\cos\frac{2\pi}{7},\quad y_5=-a\cos\frac{2\pi}{7}} \]

Explanation: \(\textbf{Propagation delays of the two sound trains:}\) The path lengths are \[ d_P=16.0\,\text{m},\qquad d_Q=24.0\,\text{m} \] Using \(v=320\,\text{m s}^{-1}\), \[ \tau_P=\frac{16.0}{320}=0.050\,\text{s} \] \[ \tau_Q=\frac{24.0}{320}=0.075\,\text{s} \] \(\textbf{Intervals during which each train reaches the microphone:}\) Both sources emit for \(0.50\,\text{s}\). Thus, P's signal arrives during \[ 0.050\lt t\lt 0.550\,\text{s} \] Q's signal arrives during \[ 0.075\lt t\lt 0.575\,\text{s} \] The overlap is the intersection of these intervals: \[ 0.075\lt t\lt 0.550\,\text{s} \] \(\textbf{Received pressure phases during overlap:}\) Each source signal has zero initial sine phase at emission. At the microphone, \[ p_P=p_0\sin[2\pi(100)(t-0.050)] \] \[ p_Q=p_0\sin[2\pi(104)(t-0.075)] \] The phase of Q relative to P, measured in cycles, is \[ \frac{\delta(t)}{2\pi} =104(t-0.075)-100(t-0.050) \] Simplifying, \[ \frac{\delta(t)}{2\pi}=4t-2.8 \] \(\textbf{Finding the first constructive-interference instant:}\) The resultant amplitude is maximum when the phase difference is an integer number of cycles. Therefore, \[ 4t-2.8=j \] where \(j\) is an integer. The earliest solution inside the overlap interval corresponds to \(j=-2\): \[ 4t-2.8=-2 \] \[ t=0.200\,\text{s} \] The next maximum corresponds to \(j=-1\): \[ 4t-2.8=-1 \] \[ t=0.450\,\text{s} \] \(\textbf{Amplitude at the intermediate observation:}\) At \(t=0.325\,\text{s}\), \[ \frac{\delta}{2\pi}=4(0.325)-2.8 \] \[ \frac{\delta}{2\pi}=-1.5 \] Thus, the waves differ by an odd multiple of \(\pi\) radians. For two equal pressure amplitudes, \[ R=2p_0\left|\cos\frac{\delta}{2}\right| \] Therefore, \[ R(0.325)=0 \] \(\textbf{Physical interpretation:}\) The sound sources start simultaneously, but their unequal path lengths delay their received wave trains by different amounts. Their different frequencies cause the relative phase to evolve during the interval of simultaneous reception. Consequently, the beat maxima depend on both emission phases and propagation delays. The microphone observes complete cancellation at the specified intermediate time because the two equal-amplitude contributions are then in opposition. Outside the overlap interval, ordinary two-wave beats cannot occur because only one train is present. \(\textbf{Final answer:}\) \[ \boxed{0.075\lt t\lt 0.550\,\text{s},\qquad t_{\max,1}=0.200\,\text{s},\quad t_{\max,2}=0.450\,\text{s},\qquad R(0.325\,\text{s})=0} \]

Explanation: The spatial displacement factor for the \(n\)th mode of a string fixed at both ends is \(\sin(n\pi x/L)\). A point-force driver placed at a displacement node of a mode cannot effectively excite that mode under the stipulated coupling model. At the midpoint, the second-harmonic spatial factor is \(\sin\pi=0\), whereas the third-harmonic factor is \(\sin(3\pi/2)\), which is nonzero. Statement I is therefore correct. At one-third of the string length, the third-harmonic spatial factor is \(\sin\pi=0\), while the second-harmonic factor is \(\sin(2\pi/3)\), which is nonzero. Statement II is correct. For the fourth harmonic, the spatial factor vanishes at the midpoint because \(\sin2\pi=0\). However, at \(x=L/3\), it is \(\sin(4\pi/3)\), which is nonzero. Thus, the fourth harmonic can be excited from the second attachment position, and Statement III is false. A resonance frequency may be allowed even when a particular point driver has no coupling to that mode. The allowed natural frequencies depend on the wave speed, length and endpoint constraints, not on whether a chosen driver happens to be placed at a node. Statement IV is false. Moving the driver changes its coupling to different spatial patterns without changing the ideal normal-mode frequencies. Statement V is correct. Therefore, the correct combination is I, II and V.

Explanation: \(\textbf{Interpreting the delay of the complete beat pattern:}\) Both sound waves propagate at the same speed in the nondispersive medium. Therefore, their combined pattern at the second microphone is a delayed copy of the pattern at the first. The microphones are separated by \[ \Delta x=15.0\,\text{m} \] The measured delay is \[ \Delta t=0.050\,\text{s} \] \(\textbf{Recovering the propagation speed:}\) \[ v=\frac{\Delta x}{\Delta t} \] Substituting, \[ v=\frac{15.0}{0.050} \] \[ v=300\,\text{m s}^{-1} \] \(\textbf{Temporal beat frequency:}\) The two frequencies are \[ f_1=500\,\text{Hz},\qquad f_2=506\,\text{Hz} \] The beat frequency is \[ f_b=|f_2-f_1| \] \[ f_b=6.0\,\text{Hz} \] \(\textbf{Temporal spacing between successive envelope maxima:}\) \[ T_b=\frac1{f_b} \] \[ T_b=\frac16\,\text{s} \] The next envelope minimum lies halfway between consecutive maxima. Therefore, \[ \Delta t_{\max\to\min}=\frac{T_b}{2} \] \[ \Delta t_{\max\to\min}=\frac1{12}\,\text{s} \] \(\textbf{Spatial variation of the relative phase:}\) For two waves travelling in the same direction, \[ \delta(x,t)=(k_2-k_1)x-(\omega_2-\omega_1)t \] Their wave-number difference has magnitude \[ |\Delta k|=\frac{2\pi f_b}{v} \] Successive spatial envelope maxima correspond to a relative-phase change of \(2\pi\). Thus, \[ |\Delta k|\Delta x_{\mathrm{beat}}=2\pi \] \[ \Delta x_{\mathrm{beat}}=\frac v{f_b} \] Substituting, \[ \Delta x_{\mathrm{beat}}=\frac{300}{6.0} \] \[ \Delta x_{\mathrm{beat}}=50\,\text{m} \] \(\textbf{Physical interpretation:}\) The beat pattern itself propagates through the common medium, so the measured delay between microphones supplies the sound speed. The beat frequency is determined by the difference between the two sound frequencies. This temporal frequency difference also produces a spatial envelope whose maxima are separated by the distance travelled during one beat period. The distance between maxima is therefore twice the distance between a maximum and its nearest minimum. The spatial minima move with time; they are not permanent stationary-wave nodes. \(\textbf{Final answer:}\) \[ \boxed{v=300\,\text{m s}^{-1},\qquad \Delta x_{\mathrm{beat}}=50\,\text{m},\qquad \Delta t_{\max\to\min}=\frac1{12}\,\text{s}} \]

Explanation: \(\textbf{Time period of pendulum P:}\) For sufficiently small angular displacement, \[ T=2\pi\sqrt{\frac Lg} \] For P, \[ T_P=2\pi\sqrt{\frac{1.00}{\pi^2}} \] \[ T_P=2.0\,\text{s} \] \(\textbf{Time period of pendulum Q:}\) The effective length of Q is \[ L_Q=1.1025\,\text{m} \] Notice that \[ 1.1025=(1.05)^2 \] Therefore, \[ T_Q=2\pi\sqrt{\frac{1.1025}{\pi^2}} \] \[ T_Q=2.10\,\text{s} \] \(\textbf{Frequencies of the pendulums:}\) \[ f_P=\frac1{T_P}=\frac12\,\text{Hz} \] For Q, \[ f_Q=\frac1{2.10} \] \[ f_Q=\frac{10}{21}\,\text{Hz} \] \(\textbf{Frequency difference in the combined optical signal:}\) Both displacement signals have equal amplitudes and begin in phase. Their superposition exhibits a slowly varying amplitude envelope. The beat frequency is \[ f_b=|f_P-f_Q| \] Substituting, \[ f_b=\frac12-\frac{10}{21} \] \[ f_b=\frac{21-20}{42} \] \[ f_b=\frac1{42}\,\text{Hz} \] \(\textbf{Beat period:}\) \[ T_b=\frac1{f_b} \] \[ T_b=42\,\text{s} \] \(\textbf{First disappearance of the resultant amplitude:}\) The individual displacements may be written as \[ x_P=A\cos(2\pi f_Pt) \] \[ x_Q=A\cos(2\pi f_Qt) \] Using the cosine addition identity, \[ x_P+x_Q=2A\cos[\pi(f_P-f_Q)t]\cos[\pi(f_P+f_Q)t] \] The amplitude envelope first vanishes when \[ \pi(f_P-f_Q)t=\frac{\pi}{2} \] Therefore, \[ t_{\mathrm{first\ zero}}=\frac1{2f_b} \] \[ t_{\mathrm{first\ zero}}=21\,\text{s} \] \(\textbf{Physical interpretation:}\) The two pendulums have slightly different lengths, so their SHM frequencies differ slightly. Their optical displacement signals therefore move into and out of phase over a much longer time than either individual oscillation period. Complete envelope cancellation occurs halfway between consecutive beat maxima because their amplitudes are equal. This optical superposition is mathematically analogous to the beats produced by two nearby sound frequencies. The individual pendulums continue oscillating even when the combined displacement signal vanishes. \(\textbf{Final answer:}\) \[ \boxed{T_P=2.0\,\text{s},\qquad T_Q=2.10\,\text{s},\qquad T_b=42\,\text{s},\qquad t_{\mathrm{first\ zero}}=21\,\text{s}} \]

Explanation: \(\textbf{Wave speed on the fixed string:}\) For transverse waves, \[ v_P=\sqrt{\frac F\mu} \] Substituting, \[ v_P=\sqrt{\frac{121}{0.010}} \] \[ v_P=110\,\text{m s}^{-1} \] \(\textbf{Frequency of the selected string mode:}\) The fourth harmonic frequency is \[ f_{4,P}=\frac{4v_P}{2L_P} \] Substituting, \[ f_{4,P}=\frac{4(110)}{2(0.50)} \] \[ f_{4,P}=440\,\text{Hz} \] The string properties remain unchanged, so this frequency stays fixed. \(\textbf{Initial fifth-harmonic frequency of the closed-open pipe:}\) For an ideal closed-open pipe, the fifth harmonic is allowed. Its frequency is \[ f_{5,Q}=\frac{5v_Q}{4L_Q} \] At \(300\,\text{K}\), \[ f_{5,Q}=\frac{5(240)}{4(0.75)} \] \[ f_{5,Q}=400\,\text{Hz} \] \(\textbf{Initial frequency difference:}\) \[ \Delta f_i=|440-400| \] \[ \Delta f_i=40\,\text{Hz} \] \(\textbf{Required sound speed for matching resonances:}\) The pipe's fifth harmonic must reach \(440\,\text{Hz}\). Therefore, \[ 440=\frac{5v_f}{4(0.75)} \] Rearranging, \[ v_f=\frac{4(0.75)(440)}5 \] \[ v_f=264\,\text{m s}^{-1} \] \(\textbf{Temperature needed for the speed change:}\) For the same ideal gas, \[ v\propto\sqrt T \] Thus, \[ \frac{T_f}{300}=\left(\frac{264}{240}\right)^2 \] \[ T_f=300(1.10)^2 \] \[ T_f=363\,\text{K} \] \(\textbf{Displacement-node counts at the matching frequencies:}\) The fourth harmonic of a fixed-fixed string contains four half-wavelength loops. Its number of interior displacement nodes is \[ N_P=4-1=3 \] The fifth harmonic of a closed-open pipe contains five quarter-wavelength segments. It has a displacement node at the closed endpoint and two additional interior displacement nodes. Thus, \[ N_Q=2 \] \(\textbf{Physical interpretation:}\) The string frequency depends on tension, linear density, length and harmonic number. The pipe frequency instead changes with the speed of sound in its air column. Heating allows the pipe's fifth harmonic to approach the unchanging fourth harmonic of the string. The two modes can have equal numerical frequencies despite their different wave types, boundary conditions and numbers of interior displacement nodes. Frequency equality does not imply that the two stationary-wave patterns are geometrically identical. \(\textbf{Final answer:}\) \[ \boxed{T_f=363\,\text{K},\qquad \Delta f_i=40\,\text{Hz},\qquad N_P=3,\quad N_Q=2} \]

Explanation: \(\textbf{Allowed frequencies for mixed endpoint conditions:}\) The fixed end is a displacement node, whereas the ideal free end is a displacement antinode. The allowed standing-wave lengths contain odd numbers of quarter wavelengths. Thus, the resonance frequencies are \[ f_h=hf_1,\qquad h=1,3,5,\ldots \] \(\textbf{Inferring the fundamental from consecutive resonances:}\) Successive allowed odd harmonics differ by twice the fundamental frequency. Therefore, \[ \Delta f=2f_1 \] From the measurements, \[ \Delta f=125-75 \] \[ \Delta f=50\,\text{Hz} \] Hence, \[ f_1=25\,\text{Hz} \] \(\textbf{Identifying the observed harmonic numbers:}\) \[ \frac{75}{25}=3,\qquad \frac{125}{25}=5,\qquad \frac{175}{25}=7 \] The observed resonances are the third, fifth and seventh harmonics. \(\textbf{Propagation speed from the mixed-boundary fundamental:}\) For a fixed-free string, \[ f_1=\frac{v}{4L} \] Therefore, \[ v=4Lf_1 \] Substituting, \[ v=4(0.60)(25) \] \[ v=60\,\text{m s}^{-1} \] \(\textbf{Determining string tension:}\) \[ v=\sqrt{\frac F\mu} \] Thus, \[ F=\mu v^2 \] \[ F=0.020(60)^2 \] \[ F=72\,\text{N} \] \(\textbf{Displacement-node positions in the fifth harmonic:}\) The \(125\,\text{Hz}\) resonance is the fifth harmonic. Its wavelength satisfies \[ L=\frac{5\lambda}{4} \] Therefore, \[ \lambda=\frac{4(0.60)}5 \] \[ \lambda=0.48\,\text{m} \] Starting from the fixed node at \(x=0\), subsequent nodes are separated by \(\lambda/2\). Thus, \[ x_1=\frac{\lambda}{2}=0.24\,\text{m} \] \[ x_2=\lambda=0.48\,\text{m} \] The right endpoint at \(x=0.60\,\text{m}\) is an antinode. \(\textbf{Physical interpretation:}\) The observed resonance spacing is twice the fundamental because mixed fixed-free boundary conditions permit only odd harmonics. This differs from the frequency spacing of a fixed-fixed string. Once the fundamental is identified, the wave speed and tension follow from the string properties. The fifth harmonic contains two interior displacement nodes, not four, because its five quarter-wavelength segments are separated by alternating nodes and antinodes. \(\textbf{Final answer:}\) \[ \boxed{f_1=25\,\text{Hz},\qquad F=72\,\text{N},\qquad x_{\mathrm{interior\ nodes}}=0.24,\ 0.48\,\text{m}} \]

Explanation: \(\textbf{Frequency difference producing the original beats:}\) \[ f_b=|f_2-f_1| \] \[ f_b=|504-500| \] \[ f_b=4.0\,\text{Hz} \] \(\textbf{Intensity relation for equal-amplitude waves with a specified relative phase:}\) Let the individual pressure amplitude of each wave be \(p_0\). Their separate time-averaged intensities are equal: \[ I_1=I_2=I_0=2.0\,\text{W m}^{-2} \] For a relative phase difference \(\delta\), the combined intensity averaged over the rapid carrier oscillations is \[ I(\delta)=I_1+I_2+2\sqrt{I_1I_2}\cos\delta \] Thus, \[ I(\delta)=4.0+4.0\cos\delta \] \(\textbf{Maximum and minimum beat-envelope intensities:}\) At maximum reinforcement, \[ \cos\delta=+1 \] Therefore, \[ I_{\max}=4.0+4.0 \] \[ I_{\max}=8.0\,\text{W m}^{-2} \] At maximum opposition, \[ \cos\delta=-1 \] Therefore, \[ I_{\min}=4.0-4.0 \] \[ I_{\min}=0 \] \(\textbf{Average intensity over one complete beat period:}\) Because the relative phase sweeps uniformly through a complete cycle during one beat period, \[ \langle\cos\delta\rangle_{\mathrm{beat}}=0 \] Hence, \[ \langle I\rangle_{\mathrm{beat}}=I_1+I_2 \] \[ \langle I\rangle_{\mathrm{beat}}=4.0\,\text{W m}^{-2} \] \(\textbf{Intensity after establishing equal frequencies:}\) After the adjustment, both sound frequencies are \(500\,\text{Hz}\). The relative phase is fixed at \[ \delta=\frac{\pi}{2} \] Thus, \[ I_f=4.0+4.0\cos\frac{\pi}{2} \] \[ I_f=4.0\,\text{W m}^{-2} \] \(\textbf{Physical interpretation:}\) Interfering sound-wave intensities cannot generally be added without accounting for the relative pressure phase. During beats, the relative phase changes continuously, causing the short-time averaged intensity to vary between complete cancellation and maximum reinforcement. When averaged over an entire beat cycle, the phase-dependent interference term vanishes. After the frequencies become equal, the relative phase is stationary and the beat modulation disappears. The chosen phase difference of \(\pi/2\) makes the new constant intensity equal to the previous beat-cycle average, although the two time histories are different. \(\textbf{Final answer:}\) \[ \boxed{f_b=4.0\,\text{Hz},\qquad I_{\max}=8.0\,\text{W m}^{-2},\quad I_{\min}=0,\qquad \langle I\rangle=I_f=4.0\,\text{W m}^{-2}} \]

Explanation: \(\textbf{Frequency of the initially excited pipe mode:}\) An ideal closed-open pipe supports odd harmonics. For its third harmonic, \[ f_3=\frac{3v}{4L} \] Initially, \[ v=320\,\text{m s}^{-1},\qquad L_i=0.80\,\text{m} \] Therefore, \[ f_{3,i}=\frac{3(320)}{4(0.80)} \] \[ f_{3,i}=300\,\text{Hz} \] The tuning fork operates at \(320\,\text{Hz}\), so the initial beat frequency is \(20\,\text{Hz}\). \(\textbf{How raising the water level changes resonance frequency:}\) The water surface acts as the closed end of the air column. Raising it shortens the column. Since \[ f_3\propto\frac1L \] the third-harmonic frequency rises as the water level rises. \(\textbf{Length at which beats first disappear:}\) Beats disappear when the pipe's excited frequency equals the fork frequency: \[ f_3=320\,\text{Hz} \] Therefore, \[ 320=\frac{3(320)}{4L_1} \] Rearranging, \[ L_1=\frac{3(320)}{4(320)} \] \[ L_1=0.75\,\text{m} \] \(\textbf{Frequency after further shortening:}\) The pipe's frequency continues increasing as the water level rises. To obtain a beat frequency of \(20\,\text{Hz}\) again, \[ |f_3-320|=20 \] The possible values are \(300\,\text{Hz}\) and \(340\,\text{Hz}\). Because the pipe is now shorter than at frequency agreement, its resonance must be above \(320\,\text{Hz}\). Thus, \[ f_3=340\,\text{Hz} \] \(\textbf{Second air-column length:}\) \[ 340=\frac{3(320)}{4L_2} \] Therefore, \[ L_2=\frac{960}{1360} \] \[ L_2=\frac{12}{17}\,\text{m} \] \(\textbf{Additional rise of the water level:}\) Raising the water level reduces air-column length. Thus, \[ \Delta h=L_1-L_2 \] Substituting, \[ \Delta h=\frac34-\frac{12}{17} \] \[ \Delta h=\frac{51-48}{68} \] \[ \Delta h=\frac3{68}\,\text{m} \] \(\textbf{Physical interpretation:}\) The tuning fork supplies a fixed reference frequency, while the pipe's resonant frequency increases as the effective air column becomes shorter. The beat rate first falls to zero when the two frequencies match. Further shortening moves the pipe frequency above the reference, making beats reappear. The second air-column length must therefore be smaller than the length at frequency agreement. The problem combines the closed-open harmonic condition with the absolute-difference nature of beat frequency. \(\textbf{Final answer:}\) \[ \boxed{L_1=0.75\,\text{m},\qquad L_2=\frac{12}{17}\,\text{m},\qquad \Delta h=\frac3{68}\,\text{m}} \]

Explanation: \(\textbf{Frequencies of the two standing-wave modes:}\) The fundamental frequency is \[ f_1=\frac{v}{2L} \] Substituting, \[ f_1=\frac{80}{2(1.00)} \] \[ f_1=40\,\text{Hz} \] Thus, the excited frequencies are \[ f_2=80\,\text{Hz},\qquad f_3=120\,\text{Hz} \] \(\textbf{Displacement at the first sensor:}\) The two contributions at \(x=L/4\) have equal positive local amplitudes and begin at positive extremes. Therefore, \[ y_2=2.0\cos(160\pi t)\quad\text{cm} \] \[ y_3=2.0\cos(240\pi t)\quad\text{cm} \] Their resultant is \[ y=2.0[\cos(160\pi t)+\cos(240\pi t)]\quad\text{cm} \] Using the cosine sum identity, \[ y=4.0\cos(200\pi t)\cos(40\pi t)\quad\text{cm} \] \(\textbf{Smallest repetition period of the actual displacement:}\) The component frequencies are \(80\) and \(120\,\text{Hz}\). Their greatest common frequency divisor is \[ \gcd(80,120)=40\,\text{Hz} \] Thus, the smallest positive period after which both components repeat their initial phases is \[ T_{\mathrm{repeat}}=\frac1{40} \] \[ T_{\mathrm{repeat}}=0.025\,\text{s} \] \(\textbf{First zero of the slowly varying amplitude envelope:}\) The nonnegative envelope is \[ R(t)=4.0|\cos(40\pi t)|\quad\text{cm} \] Its first positive zero occurs when \[ 40\pi t=\frac{\pi}{2} \] Therefore, \[ t_{\mathrm{envelope\ zero}}=\frac1{80} \] \[ t_{\mathrm{envelope\ zero}}=0.0125\,\text{s} \] The carrier cosine also passes through zero at other earlier instants, but those are not zeros of the amplitude envelope. \(\textbf{Recovering the third-mode amplitude over the full string:}\) The third-mode standing-wave displacement can be written as \[ y_3(x,t)=A_3\sin\left(\frac{3\pi x}{L}\right)\cos(240\pi t) \] At \(x=L/4\), \[ \sin\frac{3\pi}{4}=\frac1{\sqrt2} \] The local amplitude there is \(2.0\,\text{cm}\). Therefore, \[ \frac{A_3}{\sqrt2}=2.0\,\text{cm} \] \[ A_3=2\sqrt2\,\text{cm} \] \(\textbf{Evaluating the midpoint at the envelope-zero instant:}\) For the second mode, \[ \sin\left(\frac{2\pi(L/2)}{L}\right)=\sin\pi=0 \] Thus, the midpoint receives no second-mode displacement contribution. For the third mode, \[ \sin\left(\frac{3\pi(L/2)}{L}\right)=\sin\frac{3\pi}{2}=-1 \] At \(t=0.0125\,\text{s}\), \[ 240\pi t=240\pi(0.0125)=3\pi \] Therefore, \[ y(L/2,0.0125)=(2\sqrt2)(-1)\cos3\pi \] \[ y(L/2,0.0125)=+2\sqrt2\,\text{cm} \] \(\textbf{Physical interpretation:}\) The displacement at the first sensor is a sum of two harmonics of the same string. Its complete waveform repeats after the period shared by both harmonic oscillations. The amplitude envelope has a different characteristic timescale controlled by the frequency difference. At the first envelope-zero instant, the two contributions cancel at \(L/4\), but the entire string is not necessarily straight. At the midpoint, the second mode has a permanent displacement node while the third mode does not. The midpoint therefore continues oscillating and has the calculated nonzero displacement when the first sensor's contributions cancel. \(\textbf{Final answer:}\) \[ \boxed{T_{\mathrm{repeat}}=0.025\,\text{s},\qquad t_{\mathrm{first\ envelope\ zero}}=0.0125\,\text{s},\qquad y(L/2,0.0125)=+2\sqrt2\,\text{cm}} \]

Explanation: \(\textbf{Sound-speed dependence on gas composition:}\) For an ideal gas undergoing small adiabatic sound disturbances, \[ v=\sqrt{\frac{\gamma RT}{M}} \] Here \(M\) must be expressed consistently in the speed relation. The gases have equal absolute temperatures, so the common factors \(R\) and \(T\) cancel in their speed ratio. \(\textbf{Comparing the two gases:}\) \[ \frac{v_Q}{v_P} =\sqrt{\frac{\gamma_Q/M_Q}{\gamma_P/M_P}} \] Substituting the supplied values, \[ \frac{v_Q}{v_P} =\sqrt{\frac{(5/3)/4}{(7/5)/28}} \] \[ \frac{v_Q}{v_P} =\sqrt{\frac{5/12}{1/20}} \] \[ \frac{v_Q}{v_P} =\sqrt{\frac{25}{3}} \] \[ \frac{v_Q}{v_P}=\frac5{\sqrt3} \] \(\textbf{Sound speed in Q:}\) Using \(v_P=300\,\text{m s}^{-1}\), \[ v_Q=300\left(\frac5{\sqrt3}\right) \] \[ v_Q=500\sqrt3\,\text{m s}^{-1} \] \(\textbf{Comparing the wavelengths:}\) Both sound sources operate at \[ f=600\,\text{Hz} \] The travelling-wave relation is \[ \lambda=\frac vf \] Therefore, \[ \frac{\lambda_Q}{\lambda_P} =\frac{v_Q}{v_P} \] \[ \frac{\lambda_Q}{\lambda_P}=\frac5{\sqrt3} \] As a direct check, \[ \lambda_P=\frac{300}{600}=0.50\,\text{m} \] \[ \lambda_Q=\frac{500\sqrt3}{600} =\frac{5\sqrt3}{6}\,\text{m} \] \(\textbf{Effect of doubling the equilibrium pressures:}\) For an ideal gas, \[ P=\rho\frac{RT}{M} \] At unchanged temperature and composition, doubling equilibrium pressure doubles equilibrium density. The adiabatic bulk modulus is \[ B=\gamma P \] It also doubles. Consequently, \[ v'=\sqrt{\frac{2B}{2\rho}} \] \[ v'=v \] The source frequencies remain \(600\,\text{Hz}\), so neither wavelength changes. \(\textbf{Physical interpretation:}\) Sound speed depends on the elastic response relative to the inertia of the medium. Different molar masses and ratios of specific heats therefore produce different sound speeds even at the same temperature. For ideal gases at fixed temperature and composition, an isothermal increase in equilibrium pressure does not independently change sound speed. The travelling-wave wavelengths consequently remain fixed when the driving frequencies are unchanged. \(\textbf{Final answer:}\) \[ \boxed{v_Q=500\sqrt3\,\text{m s}^{-1},\qquad \frac{\lambda_Q}{\lambda_P}=\frac5{\sqrt3},\qquad \lambda'_P=\lambda_P,\quad\lambda'_Q=\lambda_Q} \]

Explanation: \(\textbf{Wave parameters from the medium and source:}\) The sound speed and frequency determine wavelength: \[ \lambda=\frac vf \] \[ \lambda=\frac{200}{100} \] \[ \lambda=2.0\,\text{m} \] Therefore, \[ k=\frac{2\pi}{\lambda}=\pi\,\text{rad m}^{-1} \] \[ \omega=2\pi f=200\pi\,\text{rad s}^{-1} \] \(\textbf{Bulk modulus of the fluid:}\) The given acoustic relation is \[ v=\sqrt{\frac B\rho} \] Hence, \[ B=\rho v^2 \] \[ B=2.0(200)^2 \] \[ B=8.0\times10^4\,\text{Pa} \] \(\textbf{Pressure disturbance from spatial compression:}\) For the longitudinal displacement, \[ s=a\sin(kx-\omega t) \] the spatial displacement gradient is \[ \frac{\partial s}{\partial x} =ak\cos(kx-\omega t) \] Therefore, \[ p'=-Bak\cos(kx-\omega t) \] At \(x=0,t=0\), the cosine factor equals one. \[ p'(0,0)=-(8.0\times10^4)(0.50\times10^{-3})(\pi) \] \[ p'(0,0)=-40\pi\,\text{Pa} \] \(\textbf{Longitudinal velocity of the fluid particle:}\) Particle velocity is the time derivative of displacement at fixed position: \[ u=\frac{\partial s}{\partial t} \] \[ u=-a\omega\cos(kx-\omega t) \] At the specified event, \[ u(0,0)=-(0.50\times10^{-3})(200\pi) \] \[ u(0,0)=-0.10\pi\,\text{m s}^{-1} \] \(\textbf{Compression or rarefaction:}\) A negative excess pressure represents pressure below the equilibrium value. Equivalently, the positive spatial displacement gradient indicates that neighbouring fluid elements are momentarily farther apart than in equilibrium. The point is therefore in a rarefaction. \(\textbf{Physical interpretation:}\) The sound disturbance travels towards positive \(x\), although the fluid particle at the observation point is moving towards negative \(x\). These directions need not agree at every phase of a longitudinal progressive wave. The sign of pressure is determined by the spatial displacement gradient rather than by the displacement value itself. At the specified event the particle displacement is zero, yet pressure and particle velocity are both nonzero. The result separates propagation direction, instantaneous particle motion and local compression state. \(\textbf{Final answer:}\) \[ p'(0,0)=-40\pi\,\text{Pa},\qquad u(0,0)=-0.10\pi\,\text{m s}^{-1},\qquad \text{rarefaction} \]

Explanation: Mechanical waves require both inertia and a restoring mechanism supplied by the medium. A liquid can resist compression through its bulk modulus, so a longitudinal pressure disturbance can propagate through it. Statement I is correct. An ideal liquid at rest does not provide the shear rigidity required for the propagation of a bulk transverse shear wave. Possessing mass and inertia alone does not supply a transverse restoring force. Statement II is therefore incorrect. A stretched string is a different mechanical arrangement because its existing tension provides the restoring force for small transverse displacements. Its particles can oscillate perpendicular to the propagation direction while the disturbance travels along the string, making Statement III correct. An elastic solid can resist compression and extension, allowing longitudinal mechanical disturbances to propagate through its bulk. Statement IV is correct. In ordinary small-amplitude mechanical wave propagation, medium particles oscillate about their equilibrium positions rather than continuously accompanying the disturbance over long distances. This distinction applies to both the transverse string pulse and the longitudinal disturbance in the solid. Statement V is therefore incorrect. The possibility of a particular mechanical-wave mode depends on the available elastic response as well as inertia. The comparison does not rule out surface waves on liquids, which involve different restoring mechanisms and are not equivalent to the bulk transverse shear wave described in Q.

Explanation: \(\textbf{Relating the source history to a travelling disturbance:}\) Let \(s(t)\) describe the displacement imposed by the source. The disturbance propagates towards positive \(x\), so the displacement at a later position is the source displacement evaluated at its retarded time: \[ y(x,t)=s\left(t-\frac{x}{v}\right) \] This relation applies after the disturbance has reached the specified position. \(\textbf{Locating the leading and trailing edges:}\) The earliest disturbance was emitted at \(t_e=0\). At the observation instant, its position is \[ x_{\mathrm{leading}}=v(1.00-0) \] \[ x_{\mathrm{leading}}=5.0\,\text{m} \] The last nonzero displacement was emitted at \(t_e=0.60\,\text{s}\). Its position is \[ x_{\mathrm{trailing}}=v(1.00-0.60) \] \[ x_{\mathrm{trailing}}=2.0\,\text{m} \] Therefore, the travelling disturbance occupies \[ 2.0\lt x\lt 5.0\,\text{m} \] \(\textbf{Position of the displacement peak:}\) The source reached its positive displacement maximum at \[ t_e=0.20\,\text{s} \] This feature travels without distortion. Its position at \(t=1.00\,\text{s}\) is \[ x_{\mathrm{peak}}=5.0(1.00-0.20) \] \[ x_{\mathrm{peak}}=4.0\,\text{m} \] \(\textbf{Particle velocity at }x=3.0\,\textbf{m:}\) The retarded time is \[ t_r=1.00-\frac{3.0}{5.0} \] \[ t_r=0.40\,\text{s} \] At the source, this time lies on the falling part of its displacement history. The displacement falls from \(+4.0\,\text{cm}\) to zero over \(0.40\,\text{s}\). Its rate of change is \[ \frac{ds}{dt}=\frac{0-4.0}{0.40} \] \[ \frac{ds}{dt}=-10\,\text{cm s}^{-1} \] Thus, the particle at \(x=3.0\,\text{m}\) is moving downwards at \[ u_y=-10\,\text{cm s}^{-1} \] \(\textbf{Particle velocity at }x=4.5\,\textbf{m:}\) Here, \[ t_r=1.00-\frac{4.5}{5.0} \] \[ t_r=0.10\,\text{s} \] This belongs to the source's rising displacement interval. The corresponding velocity is \[ u_y=\frac{4.0-0}{0.20} \] \[ u_y=+20\,\text{cm s}^{-1} \] \(\textbf{Physical interpretation:}\) A spatial snapshot of a travelling pulse records different stages of the source's earlier motion. The leading part of the pulse represents the earliest source events, while the trailing part represents later events. Consequently, the spatial pulse profile reverses the chronological ordering of the source's rise and fall. The particle velocities at two positions can therefore have opposite signs even though the entire disturbance travels in one direction. The wave speed governs propagation of the waveform, whereas the source's displacement rate determines the local transverse particle velocities. \(\textbf{Final answer:}\) \[ \boxed{2.0\lt x\lt 5.0\,\text{m},\qquad x_{\mathrm{peak}}=4.0\,\text{m},\qquad u_y(3.0)=-10\,\text{cm s}^{-1},\quad u_y(4.5)=+20\,\text{cm s}^{-1}} \]

Explanation: \(\textbf{Interpreting the sign of the pressure-velocity graph:}\) For a progressive plane sound wave travelling towards positive \(x\), excess pressure and longitudinal particle velocity have the same sign: \[ p'=\rho v u \] For a wave travelling towards negative \(x\), their signs are opposite: \[ p'=-\rho v u \] The observed graph has a negative slope, so the wave propagates towards negative \(x\). \(\textbf{Slope of the measured graph:}\) The slope is \[ m=\frac{\Delta p'}{\Delta u} \] Using the two stated points, \[ m=\frac{-45-(+90)}{(+0.10)-(-0.20)} \] \[ m=\frac{-135}{0.30} \] \[ m=-450\,\text{Pa s m}^{-1} \] \(\textbf{Recovering the wave speed:}\) For the stated left-travelling wave, \[ m=-\rho v \] Therefore, \[ v=\frac{|m|}{\rho} \] Substituting, \[ v=\frac{450}{1.5} \] \[ v=300\,\text{m s}^{-1} \] \(\textbf{Determining the bulk modulus:}\) The propagation speed in a fluid obeys \[ v=\sqrt{\frac B\rho} \] Thus, \[ B=\rho v^2 \] \[ B=1.5(300)^2 \] \[ B=1.35\times10^5\,\text{Pa} \] \(\textbf{Determining the wavelength:}\) The source frequency is \[ f=150\,\text{Hz} \] Using the travelling-wave relation, \[ \lambda=\frac vf \] \[ \lambda=\frac{300}{150} \] \[ \lambda=2.0\,\text{m} \] \(\textbf{Checking the propagation-direction inference:}\) At the recorded point with \[ p'=+90\,\text{Pa} \] the particle velocity is \[ u=-0.20\,\text{m s}^{-1} \] A positive pressure disturbance accompanied by negative particle velocity is consistent with a compression propagating towards negative \(x\). The sign relation is therefore consistent with both measured points. \(\textbf{Physical interpretation:}\) The graph slope contains more information than the magnitude of pressure or particle velocity alone. Its sign identifies propagation direction, while its magnitude gives the product of equilibrium density and sound speed. Knowing density then permits wave speed and the bulk modulus to be recovered. The independently specified frequency determines wavelength. The method relies on observing a single progressive plane wave; an arbitrary superposition of two opposite travelling waves would not generally obey the same single-slope relation. \(\textbf{Final answer:}\) \[ \text{direction }-x,\qquad v=300\,\text{m s}^{-1},\qquad B=1.35\times10^5\,\text{Pa},\qquad \lambda=2.0\,\text{m} \]

Explanation: \(\textbf{Governing wave-speed relation:}\) The wave speed on an ideal stretched string is \[ v=\sqrt{\frac{F}{\mu}} \] where \(F\) is tension and \(\mu\) is linear mass density. The initially specified string therefore has a definite propagation speed set by its mechanical properties. A change in source frequency alone does not alter that speed in the nondispersive ideal-string model. \(\textbf{Frequency and wavelength change:}\) Since \(v=f\lambda\), doubling frequency requires the wavelength to become half its original value, so Statement I is correct. When tension rises from \(64\,\text{N}\) to \(144\,\text{N}\) at fixed linear density, the speed ratio is \[ \frac{v'}v=\sqrt{\frac{144}{64}}=\frac32 \] \(\textbf{Tension and linear-density comparisons:}\) Statement II is correct. If both tension and linear density are multiplied by the same factor, their ratio remains unchanged. The wave speed therefore remains unchanged, making Statement III correct. The amplitude of a small transverse displacement affects the magnitude of string-particle motion but does not alter the ideal wave-speed relation. Doubling amplitude does not double propagation speed, so Statement IV is false. \(\textbf{Amplitude and boundary transmission:}\) At a junction between two ideal string sections driven by one steady source, the frequency of the transmitted wave remains equal to the incident frequency. Speed and wavelength can change across the junction, but frequency continuity is required by the common time dependence at the boundary. Statement V is therefore false. The correct interpretation separates source-controlled frequency and amplitude from medium-controlled propagation speed.

Explanation: \(\textbf{Inferring wave speed from the detector delay:}\) The detectors are separated by \[ \Delta x=0.15\,\text{m} \] The measured propagation time is \[ \Delta t=0.015\,\text{s} \] Therefore, \[ v=\frac{\Delta x}{\Delta t} \] \[ v=\frac{0.15}{0.015} \] \[ v=10\,\text{m s}^{-1} \] \(\textbf{Recovering wavelength:}\) The source frequency is \[ f=20\,\text{Hz} \] For a travelling wave, \[ v=f\lambda \] Hence, \[ \lambda=\frac{10}{20} \] \[ \lambda=0.50\,\text{m} \] \(\textbf{Determining the spatial phase difference:}\) The wave number is \[ k=\frac{2\pi}{\lambda} \] \[ k=4\pi\,\text{rad m}^{-1} \] The phase difference corresponding to the detector separation has magnitude \[ \Delta\phi=k\Delta x \] \[ \Delta\phi=4\pi(0.15) \] \[ \Delta\phi=\frac{3\pi}{5} \] \(\textbf{Constructing a compatible travelling-wave equation:}\) Choose the coordinate origin at detector P. Since P is at its positive displacement extreme at \(t=0\), a suitable equation is \[ y(x,t)=A\cos(kx-\omega t) \] The negative sign before \(\omega t\) represents propagation towards positive \(x\). At Q, \(x=0.15\,\text{m}\). Therefore, \[ y_Q(0)=A\cos\left(\frac{3\pi}{5}\right) \] Using the cosine identity, \[ y_Q(0)=-A\cos\left(\frac{2\pi}{5}\right) \] This is not equal to \(-A\cos(\pi/5)\). \(\textbf{Direction of Q's transverse velocity:}\) The transverse particle velocity follows from differentiation: \[ u_y=\frac{\partial y}{\partial t} \] \[ u_y=A\omega\sin(kx-\omega t) \] At Q and \(t=0\), \[ u_Q=A\omega\sin\left(\frac{3\pi}{5}\right) \] This is positive, so particle Q is moving upwards. \(\textbf{Physical interpretation:}\) The measured delay determines propagation speed because the experiment explicitly removes whole-period ambiguity. The source frequency then fixes the wavelength and spatial phase difference. At Q, the displacement is negative because its phase lies in the second quadrant. Its transverse velocity is nevertheless positive, so the particle is moving upwards towards equilibrium. This distinguishes signed displacement from instantaneous direction of particle motion. \(\textbf{Final answer:}\) \[ \boxed{v=10\,\text{m s}^{-1},\qquad\lambda=0.50\,\text{m},\qquad y_Q(0)=-A\cos\left(\frac{2\pi}{5}\right),\qquad u_Q(0)\gt 0} \]

Explanation: \(\textbf{Using the isothermal elastic modulus:}\) The speed in a fluid is \[ v=\sqrt{\frac B\rho} \] Under the student's isothermal assumption, \[ B_{\mathrm{iso}}=P \] Therefore, \[ v_{\mathrm{iso}}=\sqrt{\frac{1.0\times10^5}{1.25}} \] \[ v_{\mathrm{iso}}=\sqrt{80000} \] \[ v_{\mathrm{iso}}=200\sqrt2\,\text{m s}^{-1} \] \(\textbf{Using the appropriate adiabatic modulus:}\) For rapid, small-amplitude sound compression in the ideal gas, \[ B_{\mathrm{ad}}=\gamma P \] Substituting, \[ B_{\mathrm{ad}}=1.25(1.0\times10^5) \] \[ B_{\mathrm{ad}}=1.25\times10^5\,\text{Pa} \] Therefore, \[ v_{\mathrm{ad}}=\sqrt{\frac{1.25\times10^5}{1.25}} \] \[ v_{\mathrm{ad}}=\sqrt{100000} \] \[ v_{\mathrm{ad}}=100\sqrt{10}\,\text{m s}^{-1} \] This is equal to \(200\sqrt{5/2}\,\text{m s}^{-1}\). \(\textbf{Calculating the wavelengths:}\) The sound frequency remains \[ f=500\,\text{Hz} \] Thus, \[ \lambda_{\mathrm{iso}}=\frac{v_{\mathrm{iso}}}{f} \] \[ \lambda_{\mathrm{iso}}=\frac{200\sqrt2}{500} \] \[ \lambda_{\mathrm{iso}}=\frac{2\sqrt2}{5}\,\text{m} \] Similarly, \[ \lambda_{\mathrm{ad}}=\frac{100\sqrt{10}}{500} \] \[ \lambda_{\mathrm{ad}}=\frac{\sqrt{10}}5\,\text{m} \] \(\textbf{Fractional underestimate relative to the correct value:}\) The speed ratio is \[ \frac{v_{\mathrm{iso}}}{v_{\mathrm{ad}}} =\frac1{\sqrt\gamma} \] \[ \frac{v_{\mathrm{iso}}}{v_{\mathrm{ad}}} =\frac2{\sqrt5} \] The fractional underestimate is \[ \epsilon=\frac{v_{\mathrm{ad}}-v_{\mathrm{iso}}}{v_{\mathrm{ad}}} \] \[ \epsilon=1-\frac2{\sqrt5} \] The corresponding percentage is \[ 100\left(1-\frac2{\sqrt5}\right)\% \] \[ \epsilon_{\%}\approx10.6\% \] \(\textbf{Physical interpretation:}\) The sound-speed relation uses the appropriate elastic response of the gas during its rapid compression and expansion. For an ideal gas with \(\gamma\gt 1\), the adiabatic bulk modulus exceeds the isothermal modulus. Using the smaller isothermal modulus therefore underestimates both speed and wavelength when frequency is fixed. The fractional error must be calculated relative to the correct adiabatic speed rather than relative to the incorrect estimate. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} v_{\mathrm{iso}}=200\sqrt2\,\text{m s}^{-1},\quad v_{\mathrm{ad}}=100\sqrt{10}\,\text{m s}^{-1}\\ \lambda_{\mathrm{iso}}=\frac{2\sqrt2}{5}\,\text{m},\quad \lambda_{\mathrm{ad}}=\frac{\sqrt{10}}5\,\text{m}\\ \epsilon_{\%}=100\left(1-\frac2{\sqrt5}\right)\%\approx10.6\% \end{gathered}} \]