First slide

Applications of SHM

Question

The period of a simple pendulum is doubled, when

Moderate

A

Its length is doubled
B

The mass of the bob is doubled
C

Its length is made four times
D

The mass of the bob and the length of the pendulum are doubled

Solution

$T = 2 π \sqrt{\frac{l}{g}}$
$\Rightarrow T \propto \sqrt{l} \frac{T_{2}}{T_{1}} = \sqrt{\frac{l_{2}}{l_{1}}} 2 = \sqrt{\frac{l_{2}}{l}} l_{2} = 4 l$

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