Question

A tuning fork of frequency $512Hz$ is vibrated with a sonometer wire and $6$ beats per second are heard. The beat frequency reduces if the tension in the string is slightly increased. The original frequency of vibration of the string is

• A. $506Hz$
• B. $512Hz$
• C. $518Hz$
• D. $524Hz$

Answer 1

The correct option is A $506Hz$

Beat frequency, $f_b=\pm (f_T-f_S)$
Given $f_T=512Hz$

We have
$6=512-\frac{1}{2L}\sqrt{\frac{T}{\mu }}....\left(1\right)$
or
$6=\frac{1}{2L}\sqrt{\frac{T}{\mu }}-512....\left(2\right)$
Since, as $T$ is increased $f_b$ decreases, $\left(1\right)$ is the correct relation. Thus
$\frac{1}{2L}\sqrt{\frac{T}{\mu }}=f_S=506Hz.$