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(D) The general equation for $S.H.M.$ is $x = A \sin(\omega t + \phi)$.From the graph,the amplitude $A = 4 \, cm$.The particle passes through the mean

Vedclass · Accepted Answer

$4 \sin \left(\frac{\pi}{6} t+\frac{\pi}{6}\right)$

Vedclass · Answer

(D) The general equation for $S.H.M.$ is $x = A \sin(\omega t + \phi)$.From the graph,the amplitude $A = 4 \, cm$.The particle passes through the mean position $(x=0)$ at $t=5 \, s$ and reaches the mean position again at $t=11 \, s$. The time taken to complete half a cycle is $11 - 5 = 6 \, s$. Therefore,the time period $T = 2 	imes 6 = 12 \, s$.The angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{12} = \frac{\pi}{6} \, rad/s$.At $t = 0$,the displacement $x = 2 \, cm$. Substituting these values into the general equation:$2 = 4 \sin(\omega(0) + \phi)$$2 = 4 \sin(\phi)$$\sin(\phi) = \frac{1}{2}$$\phi = \frac{\pi}{6}$ (since the particle is moving towards the positive extreme).Thus,the equation is $x = 4 \sin \left(\frac{\pi}{6} t + \frac{\pi}{6}ight)$.

The figure shows the position-time graph of an object in $S.H.M.$ The correct equation representing this motion is ..........

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