Gujarati
14.Waves and Sound
medium

In an experiment with sonometer a tuning fork of frequency $256 Hz$ resonates with a length of $25 cm$ and another tuning fork resonates with a length of $16 cm$. Tension of the string remaining constant the frequency of the second tuning fork is ....  $Hz$

A

$163.84$

B

$400$

C

$320$

D

$204.8$

Solution

(b) In case of sonometer frequency is given by

$n = \frac{p}{{2l}}\sqrt {\frac{T}{m}} $

==> $\frac{{{n_2}}}{{{n_1}}} = \frac{{{l_1}}}{{{l_2}}} \Rightarrow {n_2} = \frac{{25}}{{16}} \times 256 = 400\,\,\,Hz$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.