First slide

Simple hormonic motion

Question

The periods of a pendulum on two planets are in the ratio 3 : 4. The acceleration due to gravity on them are in the ratio

Easy

A

9 : 16
B

3 : 4
C

4 : 3
D

16 : 9

Solution

$\large T\, = \,2\pi \sqrt {\frac{l}{g}} \, = \,T \propto \frac{1}{{\sqrt g }}$
$\large \sqrt {\frac{{{g_1}}}{{{g_2}}}} \, = \,\frac{{{T_2}}}{{{T_1}}}\, \Rightarrow \,\frac{{{g_1}}}{{{g_2}\,}}\, = \,\frac{{T_2^2}}{{T_1^2}}\, = \,\frac{{16}}{9}$

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