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) \[-\frac{\sqrt{3}}{32}{{\pi }^{2}}cm/{{s}^{2}}\]
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 \[x=a\,\sin \,\left( \frac{2\pi }{T} \right)\times t=1\,\sin \,\left( \frac{2\pi }{8} \right)t=\sin \frac{\pi }{4}t\] The velocity (v) of the particle at any instant of time 'f is \[v=\frac{dx}{dt}=\frac{d}{dt}\left[ \sin \left( \frac{\pi }{4} \right)t \right]=\frac{\pi }{4}\cos \left( \frac{\pi }{4} \right)t\] The acceleration of the particle is \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=-{{\left( \frac{\pi }{4} \right)}^{2}}\sin \left( \frac{\pi }{4} \right)t\] \[At\,t=\frac{4}{3}s\] we get \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=-{{\left( \frac{\pi }{4} \right)}^{2}}\sin \frac{\pi }{4}\times \frac{4}{3}=\frac{-\sqrt{3}{{\pi }^{2}}}{32}cm/{{s}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
