Question

. A stretched sonometer wire is in unison with a tuning fork. When the length is increased by 4%, the number of beats heard per second is 6. Find the frequency of the fork. (Ans : 156 Hz.)

Answer 1

Given :

Length = l 2 = 1.04 l 1

∴  12/ l l = 1.04

number of beats =N = 6s–1

n = ½l √(T / m )

Since l 2 > l 1 then n1 > n2  

∴ n1 – n2 = 6 Hz

For two wires of same material n1 l 1 = n2 l 2  

But , n2 = n1 – 6  

and l 2 = 1.04x l 1

∴ n1 l 1 = (n1 – 6) (1.04 xl 1 )

 n1 = (n1 – 6) (1.04)l1

 1.04 n1 – n1 = 6.24  

0.04 n1 = 6.24

n1 = 6.24/ 0.04

∴ n1 = 156 Hz

Answer 2

$\bf\huge\color{red}{ANSWER}$

Answer Credit :- @nikiswar06

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