Q. No.A cylindrical tube open at both ends has a fundamental frequency \(f_0\) in the air. The tube is dipped vertically in water such that half its length is inside water. The fundamental frequency of the air column now will be:
1. \(\frac{3f_0}{4}\)
2. \(f_0\)
3. \(\frac{f_0}{2}\)
4. \(2f_0\)
1. \(\frac{3f_0}{4}\)
2. \(f_0\)
3. \(\frac{f_0}{2}\)
4. \(2f_0\)
Subtopic: Â Standing Waves |
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A string of length \(l\) is fixed at both ends and is vibrating in second harmonic. The amplitude at antinode is \(2\) mm. The amplitude of a particle at a distance \(l/8\) from the fixed end is:
| 1. | \(2\sqrt2~\text{mm}\) | 2. | \(4~\text{mm}\) |
| 3. | \(\sqrt2~\text{mm}\) | 4. | \(2\sqrt3~\text{mm}\) |
Subtopic: Â Standing Waves |
 55%
Level 3: 35%-60%
NEET - 2022
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In an experiment with a sonometer, a tuning fork of frequency \(256~\text{Hz}\) resonates with a length of \(25~\text{cm}\) and another tuning fork resonates with a length of \(16~\text{cm}\). If the tension of the string remains constant, then the frequency of the second tuning fork will be:
1. \(163.84~\text{Hz}\)
2. \(400~\text{Hz}\)
3. \(320~\text{Hz}\)
4. \(204.8~\text{Hz}\)
1. \(163.84~\text{Hz}\)
2. \(400~\text{Hz}\)
3. \(320~\text{Hz}\)
4. \(204.8~\text{Hz}\)
Subtopic: Â Standing Waves |
 76%
Level 2: 60%+
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A string of length \(3\) m and a linear mass density of \(0.0025\) kg/m is fixed at both ends. One of its resonance frequencies is \(252\) Hz. The next higher resonance frequency is \(336\) Hz. Then the fundamental frequency will be:
1. \(84~\text{Hz}\)
2. \(63~\text{Hz}\)
3. \(126~\text{Hz}\)
4. \(168~\text{Hz}\)
Subtopic: Â Standing Waves |
 78%
Level 2: 60%+
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A string is cut into three parts, having fundamental frequencies \(n_1,n_2,\) and \(n_3\) respectively. The original fundamental frequency \(n\) is related by the expression:
1. \(\frac{1}{n}= \frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}\)
2. \(n= n_1\times n_2\times n_3\)
3. \(n= n_1+ n_2+ n_3\)
4. \(n= \frac{n_1+ n_2+ n_3}{3}\)
1. \(\frac{1}{n}= \frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}\)
2. \(n= n_1\times n_2\times n_3\)
3. \(n= n_1+ n_2+ n_3\)
4. \(n= \frac{n_1+ n_2+ n_3}{3}\)
Subtopic: Â Standing Waves |
 86%
Level 1: 80%+
AIPMT - 2000
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The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe \(L\) meter long. The length of the open pipe will be:
| 1. | \(L\) | 2. | \(2L\) |
| 3. | \(\dfrac{L}{2}\) | 4. | \(4L\) |
Subtopic: Â Standing Waves |
 79%
Level 2: 60%+
NEET - 2016
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The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is:
| 1. | \(3:1\) | 2. | \(1:2\) |
| 3. | \(2:1\) | 4. | \(1:3\) |
Subtopic: Â Standing Waves |
 63%
Level 2: 60%+
NEET - 2023
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The length of the string of a musical instrument is \(90\) cm and has a fundamental frequency of \(120\) Hz. Where should it be pressed to produce a fundamental frequency of \(180\) Hz?
| 1. | \(75\) cm | 2. | \(60\) cm |
| 3. | \(45\) cm | 4. | \(80\) cm |
Subtopic: Â Standing Waves |
 84%
Level 1: 80%+
NEET - 2020
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The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is \(20~\text{cm}\), the length of the open organ pipe is:
| 1. | \(13.2~\text{cm}\) | 2. | \(8~\text{cm}\) |
| 3. | \(12.5~\text{cm}\) | 4. | \(16~\text{cm}\) |
Subtopic: Â Standing Waves |
 64%
Level 2: 60%+
NEET - 2018
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Three waves of equal frequency having amplitudes of \(10~\mu \text{m}, 4~\mu \text{m}~\text{and}~7~\mu\text{m}\) arrive at a given point with a successive phase difference of \(\frac{\pi}{2}.\) The amplitude of the resulting wave \((\text{in}~\mu \text{m})\) is given by:
1. \(7\)
2. \(6\)
3. \(5\)
4. \(4\)
1. \(7\)
2. \(6\)
3. \(5\)
4. \(4\)
Subtopic: Â Standing Waves |
 63%
Level 2: 60%+
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