Q. No.The equation of a stationary wave is given as \(y =A\sin(0.5\pi t)\cos(0.2\pi x)\) where \(t\) is in seconds and \(x\) in centimetres. Which of the following is correct?
| 1. | Wavelength of the component waves is \(10~\text{cm}.\) |
| 2. | The separation between a node and the nearest antinode is \(2.5~\text{cm}.\) |
| 3. | Frequency of the component wave is \(0.25~\text{Hz}\). |
| 4. | All of these |
Subtopic: Â Standing Waves |
 89%
From NCERT
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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 |
 85%
From NCERT
AIPMT - 2000
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A string of length \(l\) is fixed at one end and free at the other. If it resonates in different modes, then the ratio of frequencies is:
1. \(1:2:3:~.......\)
2. \(1:3:5:7~.......\)
3. \(1:2:4:8~.......\)
4. \(1:3:9:~.......\)
1. \(1:2:3:~.......\)
2. \(1:3:5:7~.......\)
3. \(1:2:4:8~.......\)
4. \(1:3:9:~.......\)
Subtopic: Â Standing Waves |
 82%
From NCERT
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The equation of a stationary wave is \(y = 0.8\cos\left(\frac{\pi x}{20}\right)\sin200(\pi t)\), where \(x\) is in cm and \(t\) is in sec. The separation between consecutive nodes will be:
1. \(20~\text{cm}\)
2. \(10~\text{cm}\)
3. \(40~\text{cm}\)
4. \(30~\text{cm}\)
1. \(20~\text{cm}\)
2. \(10~\text{cm}\)
3. \(40~\text{cm}\)
4. \(30~\text{cm}\)
Subtopic: Â Standing Waves |
 77%
From NCERT
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An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is \(50~\text{cm}\). The next larger length of the column resonating with the same tuning fork will be:
| 1. | \(100~\text{cm}\) | 2. | \(150~\text{cm}\) |
| 3. | \(200~\text{cm}\) | 4. | \(66.7~\text{cm}\) |
Subtopic: Â Standing Waves |
 79%
From NCERT
NEET - 2016
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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 |
 76%
From NCERT
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The fundamental frequency of a closed organ pipe of a length \(20\) cm is equal to the second overtone of an organ pipe open at both ends. The length of the organ pipe open at both ends will be:
| 1. | \(80\) cm | 2. | \(100\) cm |
| 3. | \(120\) cm | 4. | \(140\) cm |
Subtopic: Â Standing Waves |
 78%
From NCERT
NEET - 2015
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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 |
 75%
From NCERT
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A string is stretched between fixed points separated by \(75.0~\text{cm}.\) It is observed to have resonant frequencies of \(420~\text{Hz}\) and \(315~\text{Hz}\). There are no other resonant frequencies between these two. The lowest resonant frequency for these strings is:
| 1. | \(155~\text{Hz}\) | 2. | \(205~\text{Hz}\) |
| 3. | \(10.5~\text{Hz}\) | 4. | \(105~\text{Hz}\) |
Subtopic: Â Standing Waves |
 76%
From NCERT
NEET - 2015
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A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio:
| 1. | \(1:2\) | 2. | \(2:3\) |
| 3. | \(3:4\) | 4. | \(4:5\) |
Subtopic: Â Standing Waves |
 73%
From NCERT
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