Question

The diameter of a wire stretched by a force of 144 N between two bridges distant 50cm is 1mm. It produces 4 beats with a tuning fork while vibrating in fundamental mode. On reducing the tension to 121 N, again 4 beats/sec are heard. The frequency of the fork is___

• A. 75 Hz
• B. 85 Hz
• C. 195 Hz
• D. 92 Hz

Answer 1

The correct option is D 92 Hz

fundamental frequency of the wire $f=1/2l\left(\sqrt{T/\mu }\right)$

so in case of 144 newton tension $f_1=12/2l\left(\sqrt{1/\mu }\right)$

and in case of 121 N tension $f_2=11/2l\left(\sqrt{1/\mu }\right)$

on reducing tension the frequency also reduces ,and as given on reducing frequency the no of beats are same, so the initial frequency of wire was 4 more than fork and final frequency of string is 4 less than fork as both time 4 beats were produced.

$f_1-f_2$ =$8$ (as described above)

$8=1/2l\left(\sqrt{1/\mu }\right)$ (difference between $f_{1}and{f}_2$ calculated from above equations )

as,

$f_1=12/2l\left(\sqrt{1/\mu }\right)$

$f_1=96$

frequecy of tuning fork will be 4 less than $f_1$

$f_{fork}=92$

correct answer is D.