First slide

Applications of SHM

Question

The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is

Easy

A

Greater than T
B

Less than T
C

Equal to T
D

Cannot be compared

Solution

Inside the mine g decreases $g_{d} = g (1 - \frac{d}{R})$
hence from $T = 2 π \sqrt{\frac{l}{g}}$ ; T increase

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