Q. No.In a stationary wave along a string, the strain is:
1. zero at the antinodes
2. maximum at the antinodes
3. zero at the nodes
4. maximum at the nodes
Subtopic: Â Standing Waves |
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The equation of a standing wave in a string is \(y = (200~\text{m})\sin\left(\frac{2\pi}{50}x\right)\cos\left(\frac{2\pi}{0.01}t\right)\) where \(x\) is in metres and \(t\) is in seconds. At the position of antinode, how many times does the distance of a string particle become \(200~\text{m}\) from its mean position in one second?
| 1. | \(100~\text{times}\) | 2. | \(50~\text{times}\) |
| 3. | \(200~\text{times}\) | 4. | \(400~\text{times}\) |
Subtopic: Â Standing Waves |
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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The maximum possible wavelength in an open organ pipe of length \(l\) is:
| 1. | \(l\) | 2. | \(2l\) |
| 3. | \(3l\) | 4. | \(4l\) |
Subtopic: Â Standing Waves |
 70%
From NCERT
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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 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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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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Stationary waves are formed on a stretched string. If the wavelength is \(\lambda,\) then the distance between two points having the maximum displacement can be:
1. \(\frac{3\lambda}{2}\)
2. \(4\lambda\)
3. \(\frac{5\lambda}{2}\)
4. All of these
1. \(\frac{3\lambda}{2}\)
2. \(4\lambda\)
3. \(\frac{5\lambda}{2}\)
4. All of these
Subtopic: Â Standing Waves |
 66%
From NCERT
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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 |
 82%
From NCERT
NEET - 2020
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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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