A) \[a\,\omega \,\cos \omega t\]
B) \[a\,\sin \omega t\]
C) \[a\,\omega \sin \omega t\]
D) \[-\frac{1}{2}(a{{\omega }^{2}}\sin \omega t)\,{{t}^{2}}\]
Correct Answer: B
Solution :
Acceleration of the particle is given as \[A=-a\,{{\omega }^{2}}\sin \omega t\] So, the velocity of the particle; \[v=\int{A\,\,dt}\] \[=\int{(-a{{\omega }^{2}}\sin \omega t)\,\,dt}\] \[=a\,\omega \cos \omega t\] \[\therefore \] Displacement of the particle, \[x=\int{v\,dt}\] \[=\int{(a\omega \cos \omega t)\,dt}\]\[a\,\sin \omega t\]
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