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 .... $Hz$
IIT 1997, Medium
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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 $=$ $8\sqrt {50} \,Hz= 56 \,Hz$
In this time number of oscillations are eight.
So time for $1$ oscillation =$\frac{1}{{8\sqrt {50} }}$
Frequency $=$ $8\sqrt {50} \,Hz= 56 \,Hz$
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