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PART - 2 CH - 13 Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 13 Oscillations2 Marks
Question
Find the length of a second pendulum on the moon where the value of gravitational acceleration is equal to the gravitational acceleration of the earth is more $\frac{1}{6}$ times. (On earth $g=9.8 m / s ^2$ )
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Answer

We know that the period of the second pendulum is 2 seconds.
$T_{m}=2 \pi \sqrt{\frac{l_{m}}{g_{m}}}$
Where $m$ has been used for the moon.
Image
Put the value :
$\begin{aligned}l_m & =\frac{(2)^2 \times\left(\frac{9.8}{6}\right)}{4 \times(3.14)^2} \\& =\frac{4 \times 9.8}{6 \times 4 \times 3.14 \times 3.14}=0.165 m\end{aligned}$

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