Q. 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. If eight oscillations are counted when the plate falls through $10\ cm,$ the frequency of the tuning fork is
A) $360\ Hz$
B) $280\ Hz$
C) $560\ Hz$
D) $56\ Hz$
Explanation
(d) Time of fall $= \ \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 10}{1000}} = \frac{1}{\sqrt{50}}$
In this time number of oscillations are eight.
So time for 1 oscillation $= \frac{1}{8\sqrt{50}}$
Frequency $= 8\sqrt{50}\ Hz = 56\ Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.