A tuning fork gives 4 beats with 50 cm length of a sonometer wire. If the length of the wire is shortened by 1 cm the number of beats is still the same. The frequency of the fork is:

Question

A tuning fork gives 4 beats with 50 cm length of a sonometer wire. If the length of the wire is shortened by 1 cm the number of beats is still the same. The frequency of the fork is: (A) 396 Hz (B) 400 Hz (C) 404 Hz (D) 384 Hz

Tardigrade Admin · Accepted Answer

According to law of length of a stretched string n propto (1/l) or, n1l1=n2l2 Given, n1=n-4,n2=n+4,l1=50 cm, l2=49 cm Hence, (n-4)× 50=(n+4)× 49 or 50 n-200=49n+196 or n=396 Hz

Q. A tuning fork gives $4$ beats with $50 c m$ length of a sonometer wire. If the length of the wire is shortened by $1 c m$ the number of beats is still the same. The frequency of the fork is:

396 Hz

400 Hz

404 Hz

384 Hz

According to law of length of a stretched string
$n \propto \frac{1}{l}$ or, $n_{1} l_{1} = n_{2} l_{2}$
Given, $n_{1} = n - 4, n_{2} = n + 4, l_{1} = 50 c m,$
$l_{2} = 49 c m$
Hence, $(n - 4) \times 50 = (n + 4) \times 49$
or $50 n - 200 = 49 n + 196$
or $n = 396 Hz$

Q. A tuning fork gives 4 beats with 50cm length of a sonometer wire. If the length of the wire is shortened by 1cm the number of beats is still the same. The frequency of the fork is: