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  • 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 g

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 (\frac{4}{3})\times1000\:\:kg/m^3 . What relationship between t and t0 is true?

Option: 1

t = t0


Option: 2

t = t0/2


Option: 3

t = 2t0


Option: 4

t = 4t0


Answers (1)

best_answer

\\ t_{0}=2 \pi \sqrt{l / g}$ \\ Due to upthrust of water on the top, its apparent weight decreases\\ upthrust = weight of liquid displaced\\ $\therefore$ Effective weight $=m g-(V \sigma g)=V \rho g-V \sigma g$\\ $V \rho g^{\prime}=V g(\rho-\sigma),$ where $\sigma$ is density of water \\ or $\quad g^{\prime}=g\left(\frac{\rho-\sigma}{\rho}\right)$\\ $t=2 \pi \sqrt{l / g^{\prime}}=2 \pi \sqrt{\frac{l \rho}{g(\rho-\sigma)}}$\\ $\therefore$ $\frac{t}{t_{0}}=\sqrt{\frac{l \rho}{g(\rho-\sigma)} \times \frac{g}{l}}=\sqrt{\frac{\rho}{\rho-\sigma}}$ $=\sqrt{\frac{4 \times 1000 / 3}{\left(\frac{4000}{3}-1000\right)}}=2$ \\ or $\quad t=t_{0} \times 2=2 t_{0}

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rishi.raj

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