A) zero
B) 2.0 m/s
C) 0.3 m/s
D) 0.4 m/s
Correct Answer: A
Solution :
For a body executing SHM the displacement equation is given by y = A \[\sin \omega \,t\] ... (i) where A is amplitude, © the angular velocity and t the time. Also, velocity \[v=\frac{dy}{dt}\] ... (ii) From Eq. (i), we get \[v=\frac{dy}{dt}=A\,\omega \,\cos \omega t\] where \[\omega =\frac{2\pi }{T}\] \[\therefore \] \[\frac{dy}{dt}=A\,.\,\frac{2\pi }{T}\cos \left( \frac{2\pi }{T}t \right)\] Given, A = 10 cm = 0.1 m, T = 4 s, \[\frac{dy}{dt}=0.1\times \frac{2\pi }{4}\cos \left( \frac{2\pi \times 1}{4} \right)\] \[\Rightarrow \] \[\frac{dy}{dt}=\frac{2\pi \times 0.1}{4}\cos \frac{\pi }{2}=0\]
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