Question
Thu March 29, 2012 By:
 

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

Expert Reply
Mon April 02, 2012
 the frequency for the both strings by nv/2l. The difference of the frequencies is equal to no of  beats.
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