First slide

Simple harmonic motion

Question

The acceleration due to gravity on the surface of the moon is $1.7 m / s^{2}$ . What is the time period of a simple pendulum on the surface of moon if it’s time period on the surface earth is 3.5 seconds?(g on the surface of earth is $9.8 m / s^{2}$ ).

Moderate

A

8.4 seconds
B

9.2 seconds
C

7.4 seconds
D

6.4 seconds

Solution

Given $g_{m} {=1.7ms}^{-2} g_{e} {=9.8ms}^{-2}$
${Time period on Earth=T}_{e} {=3.5seconds}^{} {As T}_{e} =2π \sqrt{\frac{l}{g_{e}}} T i m e p e r i o d o n m o o n = T_{m} =2π \sqrt{\frac{l}{g_{m}}} r a t i o o f a b o v e t w o e q u a t i o n s i s, \frac{T_{m}}{T_{e}} = \sqrt{\frac{g_{e}}{g_{m}}} T_{m} {=T}_{e} \sqrt{\frac{g_{e}}{g_{m}}} T_{m} =3.5 \sqrt{\frac{9.8}{1.7}} =8.4 seconds$

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