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(D) The span of oscillation $L$ is the total distance between the extreme positions,which is equal to $2A$,where $A$ is the amplitude of oscillation.T

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$2 \pi^{2} f^{2} L$

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(D) The span of oscillation $L$ is the total distance between the extreme positions,which is equal to $2A$,where $A$ is the amplitude of oscillation.Therefore,$A = \frac{L}{2}$.The angular frequency $\omega$ is given by $\omega = 2 \pi f$.The acceleration $a$ of a particle in $S.H.M.$ is given by $a = -\omega^{2} x$.At the top of the oscillation (extreme position),the displacement $x = A = \frac{L}{2}$.Substituting the values,the magnitude of acceleration is $|a| = \omega^{2} A = (2 \pi f)^{2} 	imes \frac{L}{2}$.$|a| = 4 \pi^{2} f^{2} 	imes \frac{L}{2} = 2 \pi^{2} f^{2} L$.Thus,the correct option is $D$.

$A$ man of mass $M$ is standing on a platform. The platform is executing $S.H.M.$ of frequency $f$ in the vertical direction. The span of oscillation is $L$. What is the acceleration of the platform at the top of the oscillation?

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