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Question

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 s, the tension in the string, when the bob passes through O is

A
m(g+π2gh)
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B
m(g+π2gh)
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C
m(g+π2gh)
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D
m(g+π23gh)
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Solution

The correct option is A m(g+π2gh)
Using principle of conservation of energy,
ΔKE+ΔPE=0
12mv2=mgh
v=2gh
Angular velocity of motion is ω=2πT=2π2=π

Tension in the string when bob passes through the lowest point is
T=mg+mv2r=mg+mvω (v=rω)
Substituting,
v=2gh , ω=π
we get,
T=m(g+π2gh)

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