A) The motion of the above particle is not simple harmonic
B) Simple harmonic with the period equal to that of a second's pendulum
C) Simple harmonic with the period double that of a second's a pendulum
D) Simple harmonic with amplitude \[0.4m\]
Correct Answer: B
Solution :
Given that \[Y=0.2\,\left( {{\cos }^{2}}\frac{\pi t}{2}-{{\sin }^{2}}\frac{\pi t}{2} \right)\] \[\Rightarrow \] \[Y=0.2\,\cos \pi t\] This gives, \[A=0.2,\] \[T=\frac{2\pi }{\pi }=2S\] (as \[\omega =\pi \])You need to login to perform this action.
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