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

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(D) The time $t$ taken by the plate to fall through a height $h$ is given by the equation of motion $h = \frac{1}{2}gt^2$.Given $h = 10 \, cm = 0.1 \,

Vedclass · Accepted Answer

$56$

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(D) The time $t$ taken by the plate to fall through a height $h$ is given by the equation of motion $h = \frac{1}{2}gt^2$.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 $0.1 = \frac{1}{2} 	imes 10 	imes t^2$,which gives $t^2 = 0.02$,so $t = \sqrt{0.02} \, s = \sqrt{\frac{2}{100}} \, s = \frac{\sqrt{2}}{10} \, s \approx 0.1414 \, s$.In this time,the tuning fork completes $n = 8$ oscillations.The time period of one oscillation is $T = \frac{t}{n} = \frac{\sqrt{0.02}}{8} \, s$.The frequency $f$ is the reciprocal of the time period: $f = \frac{1}{T} = \frac{8}{\sqrt{0.02}} = \frac{8}{0.1414} \approx 56.56 \, Hz$.Rounding to the nearest integer provided in the options,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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