Question

A light pointer fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating and the plate is allowed to fall freely. Eight complete oscilliations are counted when the plate falls through 10 cm, then frequency of the fork is :

• A. 65 Hz
• B. 56 Hz
• C. 46 Hz
• D. 64 Hz

Answer 1

The correct option is B 56 Hz

Given:

The displacement of the plate, $s=10cm$

Number of oscillations, $n=8$

To find:

The frequency of the fork

Solution:

From the 2nd equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

Since initial velocity is equal to zero, $u=0$

$s=\frac{1}{2}a{t}^2$

$a=g$, here the acceleration is due to gravity.

now, $t=\sqrt{\frac{2s}{g}}$

$t=\sqrt{\frac{2\left(10\right)\left({10}^{-2}\right)}{9.8}}$

$t=0.14$

But, we know, frequency is equal to number of oscillation divided by time

$\nu =\frac{n}{t}$

$\nu =\frac{8}{0.14}$

$\nu =56Hz$

The frequency of the fork is 56 Hz

Hence, the correct answer is option (B)