The bob of a simple pendulum performs SHM wit | vedclass

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Question

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.In air,the effective acceleration due to gravity is $g_{eff}

Vedclass · Accepted Answer

$T_1 = 3T$

Vedclass · Answer

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.In air,the effective acceleration due to gravity is $g_{eff} = g$.In water,the bob experiences an upward buoyant force. The effective acceleration is $g_{eff}' = g(1 - \frac{ho}{\sigma})$,where $ho$ is the density of water $(10^3 \ kg/m^3)$ and $\sigma$ is the density of the bob $(\frac{9}{8} 	imes 10^3 \ kg/m^3)$.$g_{eff}' = g(1 - \frac{10^3}{\frac{9}{8} 	imes 10^3}) = g(1 - \frac{8}{9}) = g(\frac{1}{9})$.Now,the time period in water is $T_1 = 2\pi \sqrt{\frac{l}{g_{eff}'}} = 2\pi \sqrt{\frac{l}{g/9}} = 2\pi \sqrt{\frac{9l}{g}}$.$T_1 = 3 	imes (2\pi \sqrt{\frac{l}{g}}) = 3T$.

The bob of a simple pendulum performs $SHM$ with period $T$ in air and with period $T_1$ in water. The relation between $T$ and $T_1$ is (neglect friction due to water,density of the material of the bob is $\frac{9}{8} \times 10^3 \ kg/m^3$,density of water is $1 \ g/cc$):

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