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  • The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above O and the bob is then released. Assuming the mass of

The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above O and the bob is then released. Assuming the mass of the bob to be m and time period of oscillations to be 2.0 sec, the tension in the string when the bob passes through O will be?

 

Option: 1

T=m(g+\pi \sqrt{2 g h})


Option: 2

T=m(2 g+\pi \sqrt{2 g h})


Option: 3

T=m(g+\pi \sqrt{3 g h})


Option: 4

T=m(2 g+\pi \sqrt{3 g h})


Answers (1)

best_answer

Using principle of conservation of energy,

$$ \begin{aligned} & \Delta K E+\Delta P E=0 \\ & \frac{1}{2} m v^2=m g h \\ & v=\sqrt{2 g h} \end{aligned}

\text { Angular velocity of motion is } \omega=\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi

Tension in the string when bob passes through lowest point

\begin{aligned} & T=m g+\frac{m v^2}{r}=m g+m v \omega(\because v=r \omega) \\ & \text { put } v=\sqrt{2 g h} \text { and } \omega=\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi \\ & \text { we get } T=m(g+\pi \sqrt{2 g h}) \end{aligned}

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Divya Prakash Singh

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