Q. No.1. \(\frac{3f_0}{4}\)
2. \(f_0\)
3. \(\frac{f_0}{2}\)
4. \(2f_0\)
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. \(2\sqrt3~\text{mm}\)
1. \(163.84~\text{Hz}\)
2. \(400~\text{Hz}\)
3. \(320~\text{Hz}\)
4. \(204.8~\text{Hz}\)
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. \(168~\text{Hz}\)
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. \(n= \frac{n_1+ n_2+ n_3}{3}\)
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. \(\frac{L}{2}\)
4. \(4L\)
| 1. | \(3:1\) | 2. | \(1:2\) |
| 3. | \(2:1\) | 4. | \(1:3\) |
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 |
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. \(16~\text{cm}\)
1. \(7\)
2. \(6\)
3. \(5\)
4. \(4\)
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