The displacement versus time graph of SHM is | vedclass

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(B) The displacement $y$ of a particle in $SHM$ is given by $y = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.Ac

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(B) The displacement $y$ of a particle in $SHM$ is given by $y = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.Acceleration $a$ is the second derivative of displacement with respect to time:$a = \frac{d^2y}{dt^2} = \frac{d}{dt}(A \omega \cos(\omega t)) = -A \omega^2 \sin(\omega t)$.Comparing the equations,$a = -\omega^2 y$.This means the acceleration graph is an inverted version of the displacement graph,scaled by a factor of $\omega^2$.Since the displacement graph starts at $0$ and goes positive,the acceleration graph must start at $0$ and go negative.Looking at the options,option $B$ shows a graph that starts at $0$ and goes negative,which matches the derived equation $a = -A \omega^2 \sin(\omega t)$.

The displacement versus time graph of $SHM$ is shown. Which of the following is its acceleration versus time graph?

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