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Question

(D) In simple harmonic motion $(SHM)$,the acceleration $a$ is given by $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displa

Vedclass · Accepted Answer

The acceleration of the particle is maximum at the equilibrium position.

Vedclass · Answer

(D) In simple harmonic motion $(SHM)$,the acceleration $a$ is given by $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement from the equilibrium position.At the equilibrium position,$x = 0$,so the acceleration $a = 0$,which is the minimum value.At the extreme positions,$x = \pm A$ (where $A$ is the amplitude),so the acceleration $a = \mp \omega^2 A$,which is the maximum value.Therefore,the statement that the acceleration is maximum at the equilibrium position is incorrect.The correct option is $D$.

For a particle executing simple harmonic motion,which of the following statements is not correct?

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