First slide

Applications of SHM

Question

The mass and diameter of a planet are 4 times that of the earth. What will be the time period of oscillation of a pendulum on this planet, if it is a seconds pendulum on earth

Moderate

A

$\sqrt{2}$ second
B

2 second
C

$\frac{1}{\sqrt{2}}$ second
D

4 second

Solution

$g= \frac{GM}{R^{2}}; M = mass of earth; R= radius of earth g_{planet} = \frac{{G M}_{p l a n e t}}{{R_{planet}}^{2}} g i v e n R_{p} {=4R M}_{p} =4M \frac{g_{p}}{g} = \frac{{GM}_{p}}{{R_{p}}^{2}} × \frac{R^{2}}{GM} s u b s t i t u t e g i v e n v a l u e s \frac{g_{p}}{g} = \frac{{4R}^{2}}{{16R}^{2}} = \frac{1}{4} g_{p} = \frac{g}{4} ......(1) Time period T=2π \sqrt{\frac{l}{g}} {Time period on planet=T}_{p l a n e t} =2π \sqrt{\frac{l}{g_{p l a n e t}}} d i v i d e t h e a b o v e t w o e q u a t i o n s, o n e b y o t h e r, s u b s t i t u t e e q u a t i o n (1) \frac{T}{T_{p}} = \sqrt{\frac{g_{p}}{g}} = \sqrt{\frac{g/4}{g}} = \frac{1}{\sqrt{4}} s e c o n d s p e n d u l l u m t i m e p e r i o d o n e a r t h = T = 2, s u b s t i t u t e, \frac{2}{T_{p}} = \frac{1}{\sqrt{4}} \Rightarrow T_{p} = 4 s e c o n d s$

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