First slide

Applications of SHM

Question

A pendulum that beats seconds on the surface of the earth were taken to a depth of (1/4)th the radius of the earth. What will be its time period oscillation?

Moderate

A

5.107 seconds
B

1.208 seconds
C

3.401 seconds
D

2.309 seconds

Solution

$a c c e l e r a t i o n d u e t o g r a v i t y a t a p o i n t a t a d e p t h f r o m t h e s u r f a c e o f e a r t h i s g^{1} =g (1- \frac{d}{R}), d=depth from surface of earth, R=radius of earth g i v e n d = R/4 g^{1} = g (1 - \frac{R/4}{R}) = 3 g / 4 t i m e p e r i o d o f s i m p l e p e n d u l u m i s T=2π \sqrt{\frac{l}{g}} here l=length of pendulum, g= acceleration due to gravity \Rightarrow Tα \frac{1}{\sqrt{g}} \Rightarrow T_{1} \sqrt{g_{1}} {=T}_{2} \sqrt{g_{2}} 2 \sqrt{g} = T_{2} \sqrt{\frac{3 g}{4}} 2 = T_{2} \frac{\sqrt{3}}{2} ∴ T_{2} = \frac{4}{\sqrt{3}} = 2.309 s e c o n d s$

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