Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20N force. Mass per unit length of both the strings is ame and equal to 1g/m. When both the strings vibrate simultaneously the number of beats is (1) 3 (2) 5 (3) 7 (4)8

Asked by  | 29th Mar, 2012, 12:01: AM

Expert Answer:

 the frequency for the both strings by nv/2l. The difference of the frequencies is equal to no of  beats.

Answered by  | 2nd Apr, 2012, 11:01: AM

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