First slide

Applications of SHM

Question

The length of the second pendulum on the surface of earth is 1 m. The length of seconds pendulum on the surface of moon, where g is 1/6th value of g on the surface of earth, is

Easy

A

1 / 6 m
B

6 m
C

1 / 36 m
D

36 m

Solution

$T = 2 π \sqrt{\frac{l}{g}}$ $\Rightarrow \sqrt{\frac{l}{g}}$ = constant
$\Rightarrow l \propto g;$ $\Rightarrow \frac{l_{m}}{1} = \frac{1}{6} \frac{g}{g} \Rightarrow l_{m} = \frac{1}{6} m$

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