First slide

Applications of SHM

Question

The acceleration due to gravity on moon is ${\frac{1}{6}}^{th}$ of that on the earth. If the length of the seconds pendulum is 96 cm on the earth, what is its length on the moon?

Moderate

A

16 cm
B

18 cm
C

19 cm
D

20 cm

Solution

Given $g_{m} = \frac{1}{6} g_{e} = \frac{g_{e}}{6}$
And $l_{e} = 96 cms$
For seconds pendulum time period is 2 seconds.
$T=2π \sqrt{\frac{l}{g}}$ for simple pendulum
Time remains constant in both the planets.
$\frac{l}{g} =$ constant.
$\frac{l_{m}}{l_{e}} = \frac{g_{m}}{g_{e}}$
$l_{m} = l_{e} \times \frac{g_{m}}{g_{e}}$ = $96 \times \frac{g_{e}}{\frac{6}{g_{e}}}$ $= \frac{96}{6} = 16$ cms.
length of seconds pendulum on moon = 16cm

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