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 oscillations are counted when the plate falls through 10 cm, then the frequency of the fork is (g = 9.8 m/s${}^2$)

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

Answer 1

The correct option is B 56 Hz

Time of fall $=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 10}{1000}}=\frac{1}{\sqrt{50}}$

In this the number of oscillations are eight.

So time for 1 oscillation $=\frac{1}{8\sqrt{50}}$

Frequency $=8\sqrt{50}Hz=56Hz$