First slide

Simple hormonic motion

Question

A pendulum has time period T in air. When it is made to oscillate in water, it acquired a time period $T^{'} = \sqrt{2} T$ . The density of the pendulum bob is equal to (density of water = 1)

Moderate

A

$\sqrt{2}$
B

2
C

$2 \sqrt{2}$
D

None of these

Solution

The effective acceleration of a bob in water $= g^{'} = g (1 - \frac{σ}{ρ})$ where $σ$ and $ρ$ are the densiry of water and the bob respectively. Since the period of oscillation of the bob in air and water are given as $T = 2 π \sqrt{\frac{l}{g}} and T^{'} = 2 π \sqrt{\frac{l}{g^{'}}}$
$∴ \frac{T}{T^{'}} = \sqrt{\frac{g^{'}}{g}} = \sqrt{\frac{g (1 - \frac{σ}{ρ})}{g}} = \sqrt{1 - \frac{σ}{ρ}} = \sqrt{1 - \frac{1}{ρ}}$
Putting $\frac{T}{T^{'}} = \frac{1}{\sqrt{2}} . We obtain \frac{1}{2} = 1 - \frac{1}{ρ} \Rightarrow ρ = 2$

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