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Question

The bob of a simple pendulum has a mass m and it is executing simple harmonic motion of amplitude A and period T. it collides with a body of mass m0 placed at the equilibrium position which sticks to the bob. The time period of the oscillation of the combined masses will be


A
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B
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C
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D
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Solution

The correct option is D
Let A0,T0 and ω0 respectively be the amplitude, time period and angular frequency of the combined masses. Let υ be the velocity of the bob when it reaches the equilibrium position and V0 The velocity of the combined masses just after collision. Then, from the principle of conservation of momentm, we have
mv=(m+m0)V0 (1)
Now v=r ω=r×2πT and V0 rω0=r×2πT0 where r is the length of the pendulum. Using these in (1), we get
m×(2πrT)=(m+m0)×(2πrT0)
or T0=T(m+m0m) which is choice (C).


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