First slide

Applications of SHM

Question

A simple pendulum, suspended from the ceiling of a stationary van, has time period T. If the van starts moving with a uniform velocity the period of the pendulum will be

Easy

A

Less than T
B

Equal to 2T
C

Greater than T
D

Unchanged

Solution

$T = 2 π \sqrt{\frac{l}{\sqrt{g^{2} + a^{2}}}} = 2 π \sqrt{\frac{l}{\sqrt{g^{2} + 0^{2}}}} = 2 π \sqrt{\frac{l}{g}} = T$
Effective value of ‘g’ remains unchanged.

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