- 1(current)
Q. No.The equation of vibration of a taut string, fixed at both ends, is given by; \(y=(3~\text{mm})~\text{cos}\left(\dfrac{\pi x}{10~\text{cm}}\right)~\text{sin}(800\pi~\text{s}^{-1}{t}).\) The positions of the nodes are:
| 1. | \(x= 0~\text{cm}, 10~\text{cm}, 20~\text{cm},....\) |
| 2. | \(x= 0~\text{cm}, 20~\text{cm}, 40~\text{cm},....\) |
| 3. | \(x= 5~\text{cm}, 10~\text{cm}, 15~\text{cm},....\) |
| 4. | \(x= 5~\text{cm}, 15~\text{cm}, 25~\text{cm},....\) |
Subtopic: Â Beats |
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Two different sources of sound having slightly different periods of vibration: \(1~\text{ms}\) and \(1.01~\text{ms},\) are sounded together. The resulting beat frequency is nearly:
1. \(100~\text{Hz}\)
2. \(50~\text{Hz}\)
3. \(10~\text{Hz}\)
4. \(0.01~\text{Hz}\)
1. \(100~\text{Hz}\)
2. \(50~\text{Hz}\)
3. \(10~\text{Hz}\)
4. \(0.01~\text{Hz}\)
Subtopic: Â Beats |
From NCERT
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A tuning fork, placed in a room, vibrates according to the equation: \(Y=(10^{-4}~\text m)\sin\Big({\large\frac{2\pi t}{0.01~\text s}}\Big)\) where \(Y\) is the displacement of the tip of a prong. The speed of sound in air is \(330~\text{m/s}.\)
If an additional tuning fork of frequency \(102~\text{Hz}\) is sounded together with this, then a beat frequency of:
If an additional tuning fork of frequency \(102~\text{Hz}\) is sounded together with this, then a beat frequency of:
| 1. | \(1~\text{Hz}\) will be heard. |
| 2. | \(2~\text{Hz}\) will be heard. |
| 3. | \(202~\text{Hz}\) will be heard. |
| 4. | \(101~\text{Hz}\) will be heard. |
Subtopic: Â Beats |
 86%
From NCERT
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Hints
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Q. No.
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