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(D) The equation of motion is $x=3 \sin \left(6 t+\frac{\pi}{6}\right)$,where amplitude $A=3 \ m$. At time $t=0$,the displacement is $x=3 \sin \left(\

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$1: 3$

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(D) The equation of motion is $x=3 \sin \left(6 t+\frac{\pi}{6}ight)$,where amplitude $A=3 \ m$. At time $t=0$,the displacement is $x=3 \sin \left(\frac{\pi}{6}ight) = 3 	imes \frac{1}{2} = 1.5 \ m$. The potential energy $V$ is given by $V = \frac{1}{2} k x^2$. The kinetic energy $K$ is given by $K = \frac{1}{2} k (A^2 - x^2)$. The ratio of potential energy to kinetic energy is $\frac{V}{K} = \frac{\frac{1}{2} k x^2}{\frac{1}{2} k (A^2 - x^2)} = \frac{x^2}{A^2 - x^2}$. Substituting the values $x=1.5$ and $A=3$: $\frac{V}{K} = \frac{(1.5)^2}{3^2 - (1.5)^2} = \frac{2.25}{9 - 2.25} = \frac{2.25}{6.75} = \frac{1}{3}$. Thus,the ratio is $1:3$.

The equation of motion of a particle executing simple harmonic motion is given by $x=3 \sin \left(6 t+\frac{\pi}{6}\right)$,where $x$ is in metres and $t$ is in seconds. The ratio of the potential and kinetic energies of the particle at time $t=0$ is

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