301. For an ideal pipe closed at one end and open at the other, the allowed frequencies are
ⓐ. \(f_n=\frac{nv}{2L}\), where \(n=1,2,3,\ldots\)
ⓑ. \(f_n=\frac{nv}{4L}\), where \(n=1,2,3,\ldots\)
ⓒ. \(f_n=\frac{(2n)v}{4L}\), where \(n=1,2,3,\ldots\)
ⓓ. \(f_n=\frac{(2n-1)v}{4L}\), where \(n=1,2,3,\ldots\)
Correct Answer: \(f_n=\frac{(2n-1)v}{4L}\), where \(n=1,2,3,\ldots\)
Explanation: A closed-open pipe has a displacement node at the closed end and a displacement antinode at the open end. The allowed standing-wave patterns contain an odd number of quarter-wavelength segments. Therefore, the wavelengths are \(\lambda_n=\frac{4L}{2n-1}\), where \(n=1,2,3,\ldots\). Using \(v=f_n\lambda_n\), the frequency becomes \(f_n=\frac{(2n-1)v}{4L}\). This gives the sequence \(f_1,3f_1,5f_1,\ldots\). The formula with \(n\frac{v}{2L}\) belongs to strings fixed at both ends and pipes open at both ends, not to a closed-open pipe.
302. A closed pipe of length \(0.50\,\text{m}\) contains air in which the speed of sound is \(340\,\text{m s}^{-1}\). The frequency of its third allowed mode is
ⓐ. \(170\,\text{Hz}\)
ⓑ. \(340\,\text{Hz}\)
ⓒ. \(850\,\text{Hz}\)
ⓓ. \(1020\,\text{Hz}\)
Correct Answer: \(850\,\text{Hz}\)
Explanation: \( \textbf{Given data:} \) Closed pipe length \(L=0.50\,\text{m}\), sound speed \(v=340\,\text{m s}^{-1}\).
\( \textbf{Allowed frequency relation:} \)
\[
f_n=\frac{(2n-1)v}{4L}
\]
\( \textbf{Third allowed mode:} \)
\[
n=3
\]
\( \textbf{Odd factor:} \)
\[
2n-1=5
\]
\( \textbf{Substitution:} \)
\[
f_3=\frac{5(340\,\text{m s}^{-1})}{4(0.50\,\text{m})}
\]
\( \textbf{Denominator:} \)
\[
4L=2.0\,\text{m}
\]
\( \textbf{Calculation:} \)
\[
f_3=\frac{1700}{2.0}\,\text{s}^{-1}=850\,\text{Hz}
\]
\( \textbf{Final answer:} \) The third allowed mode has frequency \(850\,\text{Hz}\).
Here “third allowed mode” means \(n=3\), which is the fifth harmonic in a closed-open pipe.
303. A pipe open at both ends and a pipe closed at one end have the same length \(L\) and contain air at the same temperature. The frequency \(3v/(4L)\) can occur as
ⓐ. the second harmonic of the open pipe only, not the closed pipe
ⓑ. the third harmonic of the closed pipe only
ⓒ. the fundamental frequency of both pipes
ⓓ. an allowed resonance in both pipes
Correct Answer: the third harmonic of the closed pipe only
Explanation: For an open-open pipe, allowed frequencies are \(f_n=\frac{nv}{2L}\), giving \(\frac{v}{2L},\frac{v}{L},\frac{3v}{2L},\ldots\). For a closed-open pipe, allowed frequencies are \(\frac{v}{4L},\frac{3v}{4L},\frac{5v}{4L},\ldots\). The frequency \(\frac{3v}{4L}\) is therefore the first overtone or third harmonic of the closed pipe. It is not in the open-pipe sequence for the same \(L\). The difference comes from the node-antinode boundary condition in the closed-open pipe. Equal pipe lengths do not imply identical harmonic series.
304. Match the air-column boundary with its displacement and pressure condition.
| Boundary | Condition |
|---|
| P. Open end | 1. Displacement node |
| Q. Closed end | 2. Displacement antinode |
| R. Open end | 3. Pressure node |
| S. Closed end | 4. Pressure antinode |
ⓐ. P-1, Q-2, R-3, S-4
ⓑ. P-2, Q-1, R-4, S-3
ⓒ. P-3, Q-4, R-2, S-1
ⓓ. P-2, Q-1, R-3, S-4
Correct Answer: P-2, Q-1, R-3, S-4
Explanation: An open end of an air column allows air particles to move most freely, so it is approximately a displacement antinode. A closed end prevents air particles from moving along the pipe at the wall, so it is a displacement node. The pressure pattern is shifted relative to the displacement pattern. At an open end, pressure variation is nearly zero, so it is a pressure node. At a closed end, pressure variation is maximum, so it is a pressure antinode. The same physical end must therefore be described differently depending on whether displacement or pressure is being considered.
305. Use the graph description below.
A standing sound wave is represented by displacement amplitude along a pipe. At the left end the amplitude is zero, and at the right end the amplitude is maximum. The pattern between the two ends is the shortest possible one.
This graph most likely represents
ⓐ. an open-open pipe in the fundamental mode
ⓑ. a closed-open pipe in the fundamental mode
ⓒ. a string fixed at both ends in the second harmonic
ⓓ. a closed-closed air column with both ends as antinodes
Correct Answer: a closed-open pipe in the fundamental mode
Explanation: The graph describes displacement amplitude, not pressure amplitude. A zero displacement amplitude at the left end indicates a displacement node, which corresponds to a closed end of a pipe. A maximum displacement amplitude at the right end indicates a displacement antinode, which corresponds to an open end. The shortest possible node-to-antinode pattern has length \(L=\frac{\lambda}{4}\). That is the fundamental mode of a closed-open pipe. An open-open pipe would have displacement antinodes at both ends, while a fixed string would have displacement nodes at both ends.
306. A closed pipe resonates at \(150\,\text{Hz}\) in its fundamental mode. The next two higher resonant frequencies are
ⓐ. \(300\,\text{Hz}\) and \(450\,\text{Hz}\)
ⓑ. \(150\,\text{Hz}\) and \(300\,\text{Hz}\)
ⓒ. \(600\,\text{Hz}\) and \(900\,\text{Hz}\)
ⓓ. \(450\,\text{Hz}\) and \(750\,\text{Hz}\)
Correct Answer: \(450\,\text{Hz}\) and \(750\,\text{Hz}\)
Explanation: \( \textbf{Given fundamental frequency:} \)
\[
f_1=150\,\text{Hz}
\]
\( \textbf{Closed-open pipe series:} \)
\[
f_1,\;3f_1,\;5f_1,\ldots
\]
\( \textbf{Next higher resonance:} \)
\[
3f_1=3(150\,\text{Hz})=450\,\text{Hz}
\]
\( \textbf{Following resonance:} \)
\[
5f_1=5(150\,\text{Hz})=750\,\text{Hz}
\]
\( \textbf{Final answer:} \) The next two higher resonant frequencies are \(450\,\text{Hz}\) and \(750\,\text{Hz}\).
The even multiples \(2f_1\) and \(4f_1\) are not allowed in an ideal closed-open pipe.
307. A resonance tube closed by water at one end resonates with a tuning fork when the shortest air-column length is \(0.17\,\text{m}\). Neglecting end correction, the wavelength of the sound is
ⓐ. \(0.17\,\text{m}\)
ⓑ. \(0.34\,\text{m}\)
ⓒ. \(0.51\,\text{m}\)
ⓓ. \(0.68\,\text{m}\)
Correct Answer: \(0.68\,\text{m}\)
Explanation: \( \textbf{Given information:} \) The tube is closed at one end by water and open at the other end.
\( \textbf{Shortest resonance condition:} \)
\[
L=\frac{\lambda}{4}
\]
\( \textbf{Given length:} \)
\[
L=0.17\,\text{m}
\]
\( \textbf{Solving for wavelength:} \)
\[
\lambda=4L
\]
\( \textbf{Substitution:} \)
\[
\lambda=4(0.17\,\text{m})
\]
\( \textbf{Calculation:} \)
\[
\lambda=0.68\,\text{m}
\]
\( \textbf{Final answer:} \) The wavelength is \(0.68\,\text{m}\).
The shortest resonance in a closed-open air column is a quarter-wavelength pattern, not a half-wavelength pattern.
308. In a resonance tube experiment with one end closed, two successive resonance lengths are \(0.18\,\text{m}\) and \(0.52\,\text{m}\). The wavelength of sound is
ⓐ. \(0.17\,\text{m}\)
ⓑ. \(0.34\,\text{m}\)
ⓒ. \(0.68\,\text{m}\)
ⓓ. \(1.04\,\text{m}\)
Correct Answer: \(0.68\,\text{m}\)
Explanation: \( \textbf{Given resonance lengths:} \)
\[
L_1=0.18\,\text{m},\quad L_2=0.52\,\text{m}
\]
\( \textbf{Successive resonance spacing in a closed-open tube:} \)
\[
L_2-L_1=\frac{\lambda}{2}
\]
\( \textbf{Difference of lengths:} \)
\[
L_2-L_1=0.52\,\text{m}-0.18\,\text{m}=0.34\,\text{m}
\]
\( \textbf{Solving for wavelength:} \)
\[
\frac{\lambda}{2}=0.34\,\text{m}
\]
\[
\lambda=0.68\,\text{m}
\]
\( \textbf{Final answer:} \) The wavelength of sound is \(0.68\,\text{m}\).
Successive resonances in a closed-open tube are separated by half a wavelength, even though the first resonance length is about a quarter wavelength.
309. A resonance curve is drawn for an air column. The vertical axis shows amplitude of vibration, and the horizontal axis shows driving frequency. A sharp peak occurs when the driving frequency is
ⓐ. much smaller than every natural frequency of the air column
ⓑ. equal to a natural frequency of the air column
ⓒ. always equal to zero
ⓓ. independent of the air-column length
Correct Answer: equal to a natural frequency of the air column
Explanation: Resonance occurs when a system is driven at one of its natural frequencies. In an air column, only certain standing-wave patterns satisfy the boundary conditions at the ends. When the driving frequency matches one of these allowed frequencies, energy transfer to the air column is especially effective. The vibration amplitude then becomes large, producing a peak in the resonance curve. The peak is not caused by zero frequency or by ignoring the length of the air column. The boundary conditions and sound speed together decide where the natural-frequency peaks occur.
310. A tuning fork of frequency \(256\,\text{Hz}\) is held above a resonance tube closed at one end. If the speed of sound is \(340\,\text{m s}^{-1}\), the approximate first resonance length, neglecting end correction, is
ⓐ. \(0.166\,\text{m}\)
ⓑ. \(0.664\,\text{m}\)
ⓒ. \(0.332\,\text{m}\)
ⓓ. \(1.328\,\text{m}\)
Correct Answer: \(0.332\,\text{m}\)
Explanation: \( \textbf{Given data:} \) Tuning fork frequency \(f=256\,\text{Hz}\), sound speed \(v=340\,\text{m s}^{-1}\).
\( \textbf{First find wavelength:} \)
\[
\lambda=\frac{v}{f}
\]
\( \textbf{Substitution:} \)
\[
\lambda=\frac{340\,\text{m s}^{-1}}{256\,\text{s}^{-1}}
\]
\( \textbf{Calculation:} \)
\[
\lambda\approx1.328\,\text{m}
\]
\( \textbf{First resonance in closed-open tube:} \)
\[
L_1=\frac{\lambda}{4}
\]
\( \textbf{Substitution:} \)
\[
L_1=\frac{1.328\,\text{m}}{4}
\]
\( \textbf{Calculation:} \)
\[
L_1\approx0.332\,\text{m}
\]
\( \textbf{Final answer:} \) The first resonance length is approximately \(0.332\,\text{m}\).
The closed end supplies a displacement node, so the first resonant length is one quarter of the wavelength.
311. If end correction \(e\) is supplied for an open end of a pipe, the effective length of a pipe open at both ends is
ⓐ. \(L_{\text{eff}}=L-e\)
ⓑ. \(L_{\text{eff}}=L+e\)
ⓒ. \(L_{\text{eff}}=L+2e\)
ⓓ. \(L_{\text{eff}}=2L+e\)
Correct Answer: \(L_{\text{eff}}=L+2e\)
Explanation: At an open end, the displacement antinode is slightly outside the physical end of the pipe. The end correction \(e\) accounts for this extra effective length. A pipe open at both ends has two open ends, so the correction is added twice. Therefore, \(L_{\text{eff}}=L+2e\). For a pipe open at one end and closed at the other, only one open end contributes an end correction. The effective length is used in the resonance formula in place of the physical length when end correction is stated.
312. An open pipe has physical length \(0.80\,\text{m}\), end correction \(0.02\,\text{m}\) at each open end, and sound speed \(336\,\text{m s}^{-1}\). Its fundamental frequency is
ⓐ. \(200\,\text{Hz}\)
ⓑ. \(210\,\text{Hz}\)
ⓒ. \(224\,\text{Hz}\)
ⓓ. \(420\,\text{Hz}\)
Correct Answer: \(200\,\text{Hz}\)
Explanation: \( \textbf{Given data:} \) Physical length \(L=0.80\,\text{m}\), end correction per open end \(e=0.02\,\text{m}\), sound speed \(v=336\,\text{m s}^{-1}\).
\( \textbf{Open-open effective length:} \)
\[
L_{\text{eff}}=L+2e
\]
\( \textbf{Substitution:} \)
\[
L_{\text{eff}}=0.80\,\text{m}+2(0.02\,\text{m})=0.84\,\text{m}
\]
\( \textbf{Fundamental condition:} \)
\[
f_1=\frac{v}{2L_{\text{eff}}}
\]
\( \textbf{Substitution:} \)
\[
f_1=\frac{336\,\text{m s}^{-1}}{2(0.84\,\text{m})}
\]
\( \textbf{Denominator:} \)
\[
2(0.84)=1.68\,\text{m}
\]
\( \textbf{Calculation:} \)
\[
f_1=200\,\text{Hz}
\]
\( \textbf{Final answer:} \) The fundamental frequency is \(200\,\text{Hz}\).
Using the physical length alone would overestimate the frequency because the effective air column is longer.
313. Two tuning forks of frequencies \(256\,\text{Hz}\) and \(260\,\text{Hz}\) are sounded together. The number of beats heard per second is
ⓐ. \(2\)
ⓑ. \(4\)
ⓒ. \(8\)
ⓓ. \(516\)
Correct Answer: \(4\)
Explanation: \( \textbf{Given frequencies:} \)
\[
f_1=256\,\text{Hz},\quad f_2=260\,\text{Hz}
\]
\( \textbf{Beat frequency relation:} \)
\[
f_b=|f_1-f_2|
\]
This relation applies when two sounds of nearly equal frequencies are heard together.
\( \textbf{Substitution:} \)
\[
f_b=|256-260|
\]
\( \textbf{Calculation:} \)
\[
f_b=4\,\text{Hz}
\]
\( \textbf{Meaning:} \) \(4\,\text{Hz}\) beat frequency means \(4\) beats per second.
\( \textbf{Final answer:} \) The number of beats heard per second is \(4\).
The beat frequency is the difference of the two frequencies, not their sum or average.
314. Beats are most clearly heard when two sound waves reaching the ear have
ⓐ. exactly the same frequency and no amplitude
ⓑ. very widely different frequencies with no steady beat pattern
ⓒ. nearly equal frequencies and comparable amplitudes
ⓓ. frequencies that are both zero
Correct Answer: nearly equal frequencies and comparable amplitudes
Explanation: Beats are periodic variations in loudness caused by superposition of two waves of slightly different frequencies. The waves alternately reinforce and weaken each other as their phase difference changes slowly. If the frequencies are nearly equal, this loudness variation is slow enough to be heard as beats. If the frequencies are very far apart, the variation is too rapid to be perceived as ordinary beats. Comparable amplitudes make the loudness changes more noticeable. Beats therefore depend on a small frequency difference, not on a large separation of frequencies.
315. Use the graph description below.
A graph of resultant sound amplitude against time shows rapid oscillations inside a slowly varying envelope. The envelope has repeated maxima separated by \(0.25\,\text{s}\).
The beat frequency is
ⓐ. \(0.25\,\text{Hz}\)
ⓑ. \(2\,\text{Hz}\)
ⓒ. \(4\,\text{Hz}\)
ⓓ. \(8\,\text{Hz}\)
Correct Answer: \(4\,\text{Hz}\)
Explanation: \( \textbf{Graph information:} \) Successive loudness or envelope maxima are separated by \(0.25\,\text{s}\).
\( \textbf{Beat period:} \)
\[
T_b=0.25\,\text{s}
\]
\( \textbf{Beat frequency relation:} \)
\[
f_b=\frac{1}{T_b}
\]
\( \textbf{Substitution:} \)
\[
f_b=\frac{1}{0.25\,\text{s}}
\]
\( \textbf{Calculation:} \)
\[
f_b=4\,\text{s}^{-1}=4\,\text{Hz}
\]
\( \textbf{Final answer:} \) The beat frequency is \(4\,\text{Hz}\).
The rapid oscillations correspond to the sound waves themselves, while the slow envelope gives the beat period.
316. Two tuning forks produce \(5\) beats per second. One fork has frequency \(300\,\text{Hz}\). Without any extra information, the other fork may have frequency
ⓐ. \(295\,\text{Hz}\) or \(305\,\text{Hz}\)
ⓑ. \(150\,\text{Hz}\) or \(600\,\text{Hz}\)
ⓒ. \(5\,\text{Hz}\) only
ⓓ. \(300\,\text{Hz}\) only
Correct Answer: \(295\,\text{Hz}\) or \(305\,\text{Hz}\)
Explanation: Beats per second give the beat frequency \(f_b\). Here \(f_b=5\,\text{Hz}\). The beat relation is \(f_b=|f_1-f_2|\). If one fork has frequency \(300\,\text{Hz}\), the other frequency must differ from it by \(5\,\text{Hz}\). Therefore, the possible values are \(300-5=295\,\text{Hz}\) and \(300+5=305\,\text{Hz}\). Beat data alone gives the size of the difference, not whether the unknown fork is higher or lower in frequency. Extra tuning information is needed to remove this ambiguity.
317. A tuning fork of unknown frequency produces \(6\) beats per second with a \(256\,\text{Hz}\) fork. When a little wax is attached to the unknown fork, the beat frequency decreases to \(4\) beats per second. The original unknown frequency was
ⓐ. \(250\,\text{Hz}\)
ⓑ. \(252\,\text{Hz}\)
ⓒ. \(260\,\text{Hz}\)
ⓓ. \(262\,\text{Hz}\)
Correct Answer: \(262\,\text{Hz}\)
Explanation: \( \textbf{Initial beat frequency:} \)
\[
|f-256|=6
\]
\( \textbf{Initial possible frequencies:} \)
\[
f=250\,\text{Hz}\quad \text{or}\quad f=262\,\text{Hz}
\]
\( \textbf{Effect of attaching wax:} \) Wax increases the effective mass of the fork and lowers its frequency.
\( \textbf{Test if \(f=250\,\text{Hz}\):} \) Lowering the unknown frequency below \(250\,\text{Hz}\) would make its difference from \(256\,\text{Hz}\) larger than \(6\,\text{Hz}\), not \(4\,\text{Hz}\).
\( \textbf{Test if \(f=262\,\text{Hz}\):} \) Lowering the unknown frequency can bring it closer to \(256\,\text{Hz}\), reducing the beat frequency from \(6\,\text{Hz}\) to \(4\,\text{Hz}\).
\( \textbf{Final answer:} \) The original unknown frequency was \(262\,\text{Hz}\).
The change after loading tells whether the unknown fork started above or below \(256\,\text{Hz}\).
318. For two waves of frequencies \(f_1\) and \(f_2\) producing beats, the average pitch heard is mainly associated with
ⓐ. \(\frac{f_1+f_2}{2}\)
ⓑ. \(\frac{|f_1-f_2|}{2}\)
ⓒ. \(f_1f_2\)
ⓓ. \(\frac{f_1}{f_2}\)
Correct Answer: \(\frac{f_1+f_2}{2}\)
Explanation: When two close frequencies are superposed, the sound amplitude rises and falls with the beat frequency \(|f_1-f_2|\). The rapid oscillation inside the beat envelope has a frequency close to the average of the two frequencies. That average frequency is \(\frac{f_1+f_2}{2}\). The ear usually associates the pitch with this rapid oscillation, while the slow variation is heard as beats. The difference frequency controls the loudness variation, not the central pitch. This separates the beat frequency from the average frequency of the sound.
319. Two tuning forks of frequencies \(440\,\text{Hz}\) and \(444\,\text{Hz}\) are sounded together. During the beat formation, the loudness becomes maximum
ⓐ. \(8\) times per second
ⓑ. \(442\) times per second
ⓒ. \(884\) times per second
ⓓ. \(4\) times per second
Correct Answer: \(4\) times per second
Explanation: \( \textbf{Given frequencies:} \)
\[
f_1=440\,\text{Hz},\quad f_2=444\,\text{Hz}
\]
\( \textbf{Beat frequency relation:} \)
\[
f_b=|f_2-f_1|
\]
\( \textbf{Substitution:} \)
\[
f_b=|444-440|\,\text{Hz}
\]
\( \textbf{Calculation:} \)
\[
f_b=4\,\text{Hz}
\]
\( \textbf{Meaning of beat frequency:} \) \(4\,\text{Hz}\) means the loudness rises to a maximum \(4\) times per second.
\( \textbf{Final answer:} \) The loudness becomes maximum \(4\) times per second.
The average frequency gives the pitch region, while the difference frequency gives the slow loudness variation.
320. Two waves reaching a point are represented by \(y_1=A\sin\omega_1t\) and \(y_2=A\sin\omega_2t\), where \(\omega_1\) and \(\omega_2\) are close. The resultant displacement can be written in the form
ⓐ. \(2A\cos\left(\frac{\omega_1-\omega_2}{2}t\right)\sin\left(\frac{\omega_1+\omega_2}{2}t\right)\)
ⓑ. \(2A\sin\left(\frac{\omega_1-\omega_2}{2}t\right)\cos\left(\frac{\omega_1+\omega_2}{2}t\right)\)
ⓒ. \(A\sin[(\omega_1+\omega_2)t]\)
ⓓ. \(A\cos[(\omega_1-\omega_2)t]\)
Correct Answer: \(2A\cos\left(\frac{\omega_1-\omega_2}{2}t\right)\sin\left(\frac{\omega_1+\omega_2}{2}t\right)\)
Explanation: \( \textbf{Given waves:} \)
\[
y_1=A\sin\omega_1t,\quad y_2=A\sin\omega_2t
\]
\( \textbf{Resultant by superposition:} \)
\[
y=y_1+y_2=A\sin\omega_1t+A\sin\omega_2t
\]
\( \textbf{Trigonometric identity used:} \)
\[
\sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)
\]
\( \textbf{Apply the identity:} \)
\[
y=2A\sin\left(\frac{\omega_1+\omega_2}{2}t\right)\cos\left(\frac{\omega_1-\omega_2}{2}t\right)
\]
\( \textbf{Same product written in option order:} \)
\[
y=2A\cos\left(\frac{\omega_1-\omega_2}{2}t\right)\sin\left(\frac{\omega_1+\omega_2}{2}t\right)
\]
\( \textbf{Final answer:} \) The resultant has a fast oscillation multiplied by a slowly varying amplitude factor.
The cosine factor controls the beat envelope, while the sine factor carries the average-frequency oscillation.
