Wave Mechanics Question 557
Question: The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4 / 3 s is
Options:
A) $ \frac{\sqrt{3}}{32}{{\pi }^{2}}cm/s^{2} $
B) $ \frac{-{{\pi }^{2}}}{32}cm/s^{2} $
C) $ \frac{{{\pi }^{2}}}{32}cm/s^{2} $
D) $ -\frac{\sqrt{3}}{32}{{\pi }^{2}}cm/s^{2} $
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Answer:
Correct Answer: D
Solution:
[d] From the graph it is clear that the amplitude is 1 cm and the time period is 8 second.
Therefore the equation for the S.H.M. is $ x=a,\sin ,( \frac{2\pi }{T} )\times t=1,\sin ,( \frac{2\pi }{8} )t=\sin \frac{\pi }{4}t $ The velocity (v) of the particle at any instant of time ‘f is $ v=\frac{dx}{dt}=\frac{d}{dt}[ \sin ( \frac{\pi }{4} )t ]=\frac{\pi }{4}\cos ( \frac{\pi }{4} )t $ The acceleration of the particle is $ \frac{d^{2}x}{dt^{2}}=-{{( \frac{\pi }{4} )}^{2}}\sin ( \frac{\pi }{4} )t $
$ At,t=\frac{4}{3}s $ we get $ \frac{d^{2}x}{dt^{2}}=-{{( \frac{\pi }{4} )}^{2}}\sin \frac{\pi }{4}\times \frac{4}{3}=\frac{-\sqrt{3}{{\pi }^{2}}}{32}cm/s^{2} $
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