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Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is
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- 7
- 8
- 3
- 5
Correct Option: A
The frequency of vibration of a string is given by,
| f = | √ | ||
| 2I | m |
where m is mass per unit length.
| f1 = | √ | f2 = | √ | ||||
| 2I1 | m | 2I2 | m |
| f2 - f1 = | √ | ||||
| 2 | m | l1l2 |
| √ | = √ | = √2 10 × 10² = 1.414 × 100 | ||
| m | 10-3 |
= 141.4
| = | |||
| l1l2 | 51.6 × 49.1 |
| = | = | ||
| 50 × 50 | 10 |
| ∴ f2 - f1 = | × 141.4 × | = 7 beats | ||
| 2 | 10 |
