Question

For a particle starting from rest, the acceleration varies with time according to relation, $ A=-a{{\omega }^{2}}\sin \omega t $ The displacement of this particle at a time t will be

• 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}} $

Answer 1

Answer: (B)

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 $