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Question 47 of 66

Q.The amplitude and time period of a simple pendulum bob are 0.05 m and 2 s respectively. Then the maximum velocity of the bob is :

(a) 0.157 ms^-1
(b) 0.257 ms^-1
(c) 0.10 ms^-1
(d) 0.025 ms^-1
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2019MCQ· 1mImportance★★★★★
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The maximum velocity of a simple pendulum bob is v_max = Aomega = A(2*pi/T) = 0.157 m/s for A = 0.05 m and T = 2 s.

A simple pendulum executes simple harmonic motion, described by x = A sin(omega t), where A is the amplitude and omega = 2pi/T is the angular frequency. The velocity is v = dx/dt = Aomega*cos(omega t)

This is maximum when cos(omega t) = 1, i.e.

v_max = A*omega

…

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