For a particle performing Simple Harmonic Mot | vedclass

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(A) For a particle performing $S.H.M.$,the displacement $x$ as a function of time $t$ is given by $x = A \sin(\omega t)$,where $A$ is the amplitude an

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(A) For a particle performing $S.H.M.$,the displacement $x$ as a function of time $t$ is given by $x = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.According to Hooke's Law for $S.H.M.$,the restoring force $F$ is given by $F = -kx$,where $k$ is the force constant.Substituting the expression for $x$,we get $F = -k(A \sin(\omega t)) = -kA \sin(\omega t)$.This equation shows that the force $F$ is proportional to the negative of the displacement $x$.Therefore,the force-time graph will be an inverted version of the displacement-time graph. If the displacement graph starts from the origin and goes positive,the force graph must start from the origin and go negative.

For a particle performing Simple Harmonic Motion $(S.H.M.)$,the displacement-time graph is as shown. For that particle,the force-time graph is correctly represented by which of the following graphs?

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