First slide

Applications of SHM

Question

On a planet a body is left fall freely from a height 2 metres reaches the ground $\sqrt{2}$ seconds if the length of a simple pendulum is 2 metres on that planet, its time period is

Moderate

A

3.14 seconds
B

6.28 seconds
C

1.57 seconds
D

4.17 seconds

Solution

$h= \frac{1}{2} {gt}^{2}$ given length of the pendulum l=2m, give time of fall t= $\sqrt{2}$ sec
$2= \frac{1}{2} g_{p} {(\sqrt{2})}^{2}$
$g_{p} = \frac{4}{2} {=2 m/sec}^{2}$
Time period $T=2π \sqrt{\frac{l}{g}} =2π \sqrt{\frac{2}{2}} =2π=2(3.14)$
$T=6.28$ seconds

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