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

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$90$

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(B) The displacement of a particle executing $SHM$ is given by $x(t) = A \sin(\omega t + \phi)$.The velocity of the particle is the time derivative of displacement: $v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.We can rewrite the velocity expression using the trigonometric identity $\cos(	heta) = \sin(	heta + 90^{\circ})$:$v(t) = A \omega \sin(\omega t + \phi + 90^{\circ})$.Comparing the phase of displacement $(\omega t + \phi)$ and velocity $(\omega t + \phi + 90^{\circ})$,the phase difference is $90^{\circ}$ or $\frac{\pi}{2} 	ext{ radians}$.

What is the phase difference between the velocity and displacement of a particle executing $SHM$ (in $^{\circ}$)?

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