A pendulum has time period T in air. When it is made to oscillate in water, it acquired a time period T'$$=2T$$. The specific gravity of the pendulum bob is equal to :

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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 specific gravity of the pendulum bob is equal to :

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Solution

$\begin{array}{l} The effective acceleration of a bob in water = g^{'} = g (1 - \frac{d}{D}), \\ where d and D are the density of water and the bob respectively. \\ \frac{D}{d} = specific gravity of the bob. \\ Since the period of oscillation of the bob in air and water are given as \\ T = 2 π \sqrt{\frac{ℓ}{g}} and T' = 2 π \sqrt{\frac{ℓ}{g^{'}}} \\ ∴ T / T^{'} = \sqrt{\frac{g^{'}}{g}} = \sqrt{\frac{g (1 - d / D)}{g}} = \sqrt{1 - \frac{d}{D}} = \sqrt{1 - \frac{1}{s}} \\ Putting T / T^{'} = 1 / \sqrt{2}, we obtain \\ \Rightarrow \frac{1}{2} = 1 - \frac{1}{s} \\ \Rightarrow \frac{1}{s} = \frac{1}{2} \\ \Rightarrow s = 2 \end{array}$

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Applications of SHM

Remember concepts with our Masterclasses.

A pendulum has time period T in air. When it is made to oscillate in water, it acquired a time period T'=2T. The specific gravity of the pendulum bob is equal to :

Ready to Test Your Skills?

free study material

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 specific gravity of the pendulum bob is equal to :