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(B) The displacement $(x)$ of a particle in $SHM$ is given by $x = A \sin(\omega t + \phi)$.The velocity $(v)$ is the first derivative of displacement

Vedclass · Accepted Answer

$\pi \; rad$

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(B) The displacement $(x)$ of a particle in $SHM$ is given by $x = A \sin(\omega t + \phi)$.The velocity $(v)$ is the first derivative of displacement: $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.The acceleration $(a)$ is the second derivative of displacement: $a = \frac{d^2x}{dt^2} = -A \omega^2 \sin(\omega t + \phi)$.Using the trigonometric identity $-\sin(	heta) = \sin(	heta + \pi)$,we can rewrite the acceleration as $a = A \omega^2 \sin(\omega t + \phi + \pi)$.Comparing the phase of displacement $(\omega t + \phi)$ and acceleration $(\omega t + \phi + \pi)$,the phase difference is $\pi \; rad$.

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is

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