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13.Oscillations
hard
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 $O$ is
A
$m\,(g + \pi \sqrt {2g\,h} )$
B
$m\,(g + \sqrt {{\pi ^2}g\,h} )$
C
$m\,\left( {g + \sqrt {\frac{{{\pi ^2}}}{2}g\,h} } \right)$
D
$m\,\left( {g + \sqrt {\frac{{{\pi ^2}}}{3}g\,h} } \right)$
Solution
(a) Tension in the string when bob passes through lowest point
$T = mg + \frac{{m{v^2}}}{r} = mg + mv\omega $ ( $v = r\omega$)
putting $v = \sqrt {2gh} $ and $\omega= \frac{{2\pi }}{T} = \frac{{2\pi }}{2} = \pi $
we get $T = m\;(g + \pi \sqrt {2gh} )$
Standard 11
Physics
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