The displacement-time graph of a particle exe | vedclass

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(D) For a particle executing $S.H.M.$,the displacement $y$ is given by $y = A \sin(\omega t)$.Acceleration $a = \frac{d^2y}{dt^2} = -A\omega^2 \sin(\o

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(D) For a particle executing $S.H.M.$,the displacement $y$ is given by $y = A \sin(\omega t)$.Acceleration $a = \frac{d^2y}{dt^2} = -A\omega^2 \sin(\omega t) = -\omega^2 y$.According to Newton's second law,the restoring force $F$ is given by $F = ma = -m\omega^2 y$.Substituting the expression for $y$,we get $F = -m\omega^2 A \sin(\omega t)$.This shows that the force $F$ is also a sinusoidal function of time but with a phase difference of $\pi$ radians relative to the displacement $y$.Therefore,if the displacement graph is a positive sine wave starting from the origin,the force-time graph must be an inverted (negative) sine wave starting from the origin,which corresponds to option $D$.

The displacement-time graph of a particle executing $S.H.M.$ is as shown in the figure. The corresponding force-time graph of the particle is:

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