A light pointer fixed to one prong of a tunin | vedclass

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(D) The time taken for the plate to fall through a distance $h$ under gravity is given by $h = \frac{1}{2} g t^2$.Given $h = 10 \, cm = 0.1 \, m$ and

Vedclass · Accepted Answer

$56$

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(D) The time taken for the plate to fall through a distance $h$ under gravity is given by $h = \frac{1}{2} g t^2$.Given $h = 10 \, cm = 0.1 \, m$ and $g = 9.8 \, m/s^2$ (or $10 \, m/s^2$ for simplicity).Using $g = 10 \, m/s^2$,we have $0.1 = \frac{1}{2} 	imes 10 	imes t^2$.$0.1 = 5 t^2 \implies t^2 = 0.02 \implies t = \sqrt{0.02} = \sqrt{\frac{2}{100}} = \frac{\sqrt{2}}{10} \, s$.In this time $t$,the tuning fork completes $n = 8$ oscillations.The frequency $f$ is given by $f = \frac{n}{t}$.$f = \frac{8}{\frac{\sqrt{2}}{10}} = \frac{80}{\sqrt{2}} = 40 \sqrt{2} \approx 40 	imes 1.414 = 56.56 \, Hz$.Rounding to the nearest provided option,the frequency is $56 \, Hz$.

$A$ 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 .... $Hz$

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