Waves - Result Question 58

60. Each of the two strings of length 51.6cm and 49.1cm are tensioned separately by 20N force. Mass per unit length of both the strings is same and equal to 1g/m. When both the strings vibrate simultaneously the number of beats is

[2009]

(a) 7

(b) 8

(c) 3

(d) 5

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Answer:

Correct Answer: 60. (a)

Solution:

  1. (a) The frequency of vibration of a string is given by, f=121Tm where m is mass per unit length.

f1=12l1Tm,f2=12l2Tm,

f2f1=12Tm(l1l2)l1l2

Tm=20103=2×102=1.414×100

=141.4

111211l2=(51.649.1)×10251.6×49.1

=2.5×10250×50=110

f2f1=12×141.4×110=7 beats



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