Each of the two strings of length 51.6 cm and 49.1 cm are tensed 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 ......

Each of the two strings of length 51.6 cm and 49.1 cm are tensed separ

1.	Each of the two strings of length 51.6 cm and 49.1 cm are tensed 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 ......
Answer» 7 8 3 5 Solution :Here FREQUENCY of VIBRATION of string is, `f =(1)/( 2L ) sqrt ((T)/( MU ))` `therefore f _(1) = (1)/( 2 (0.516)) sqrt ((20)/( 0.001)) = 144 Hz` `therefore` Beat frequency `f _(2) - f _(1) = 144 - 137 =7 Hz`