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Question

(A) In $S.H.M.$,the displacement of a particle is given by $x(t) = A \cos(\omega t + \phi)$.Since the particle starts from the extreme position,at $t

Vedclass · Accepted Answer

$\pi \ rad$

Vedclass · Answer

(A) In $S.H.M.$,the displacement of a particle is given by $x(t) = A \cos(\omega t + \phi)$.Since the particle starts from the extreme position,at $t = 0$,$x = A$,which implies $\phi = 0$. Thus,$x(t) = A \cos(\omega t)$.The acceleration $a(t)$ is given by the second derivative of displacement: $a(t) = \frac{d^2x}{dt^2} = -\omega^2 A \cos(\omega t)$.We can rewrite the acceleration as $a(t) = \omega^2 A \cos(\omega t + \pi)$.Comparing the phase of displacement $(\omega t)$ and acceleration $(\omega t + \pi)$,the phase difference is $\pi \ rad$.

$A$ particle is performing $S.H.M.$ starting from the extreme position. The graphical representation shows that between displacement and acceleration,there is a phase difference of:

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