The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t₀ in air. Neglecting frictional force of water and given that the density of the bob is (4/3) × 1000 kg/m³. What relationship between t and t₀ is true (a) t = 2t₀ (b) t = t₀/2 (c) t = t₀ (d) t = 4t₀

Answer: (a). geff = g(1 - (ρw)/(ρb)) = g(1 - (1000)/(4000/3)) = g(1-3/4) = g/4. t = t₀√ggeff = t₀√(4). So the answer is (a) t = 2t₀.

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Simple Pendulum

The bob of a simple pendulum executes simple harmonic motion in water with a period $t,$ while the period of oscillation of the bob is $t_0$ in air. Neglecting frictional force of water and given that the density of the bob is $(4/3) \times 1000\text{ kg/m}^{3}.$ What relationship between $t$ and $t_0$ is true

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Answer: (a)

$g_{eff} = g\left(1 - \frac{\rho_w}{\rho_b}\right) = g\left(1 - \frac{1000}{4000/3}\right) = g\left(1-\frac{3}{4}\right) = \frac{g}{4}$ .
$t = t_0\sqrt{\frac{g}{g_{eff}}} = t_0\sqrt{4}$ .
So the answer is (a) $t = 2t_0$ .

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