A) Along a circular path
B) Simple harmonic motion with time period 0.1 s
C) No simple harmonic motion
D) Simple harmonic motion with time period 0.2 s
Correct Answer: B
Solution :
The displacement equation is given by \[y=0.2\,\sin \left( 10\pi \,t+1.5\pi \right)\,\cos \,\left( 10\,\pi \,t+1.5\pi \right)\] Rearranging the terms, we have \[y=0.1\left[ 2\,\sin \,\left( 10\,\pi t+1.5\,\pi \right)\cos \,\left( 10\,\pi t+1.5\pi \right) \right]\] Using 2 sin A cos A = sin 2 A, we have \[y=0.1\,\,\sin \left( 20\,\pi \,t+3\pi \right)\] ?(1) This equation takes the form of displacement equation in SHM, we have \[y=a\,\sin \left( \omega \,t+\phi \right)\] ?(2) Where a is amplitude, \[\omega \] is angular velocity, \[\phi \] is phase difference. Comparing Eqs. (1) and (2), we get \[\omega t=20\,\pi \,t\] \[\Rightarrow \] \[\omega =20\,\pi \] Also time-period=\[\frac{2\pi }{\omega }=\frac{2\,\pi }{20\,\pi }=0.1\,s\]
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