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 is R. What will be its time period of oscillation?

Moderate

A

5.107 seconds
B

1.208 seconds
C

3.401 seconds
D

2.309 seconds

Solution

$f r o m g r a v i t a t i o n, a c c e l e r a t i o n d u e t o g r a v i t y a t a d e p t h d f r o m s u r f a c e o f e a r t h i s, g^{1} =g (1- \frac{d}{R}), here given d= \frac{R}{4}, substitute g^{1} = g (1 - \frac{R/4}{R}) = 3 g / 4 = g_{2} (s u p p o s e) t i m e p e r i o d o f s i m p l e p e n d u l u m i s T=2π \sqrt{\frac{l}{g}} \Rightarrow Tα \frac{1}{\sqrt{g}}, where length l of pendulum is constant \Rightarrow T \sqrt{g} = c o n s \tan t T_{1} \sqrt{g_{1}} {=T}_{2} \sqrt{g_{2}} {substitute g}_{1} {=g and g}_{2} = \frac{3 g}{4} p e n d u l u m b e a t s s e c o n d s m e a n s, T_{1} =2 seconds 2 \sqrt{g} = T_{2} \sqrt{\frac{3 g}{4}} 2 = T_{2} \frac{\sqrt{3}}{2} ∴ T_{2} = \frac{4}{\sqrt{3}} = 2.309 s e c o n d s$

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