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(C) The total energy of a particle in simple harmonic motion is equal to its maximum potential energy,which is given by $E = \frac{1}{2} k A^2 = 25 \

Vedclass · Accepted Answer

$18.75$

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(C) The total energy of a particle in simple harmonic motion is equal to its maximum potential energy,which is given by $E = \frac{1}{2} k A^2 = 25 \ J$.The potential energy at any displacement $x$ is given by $U = \frac{1}{2} k x^2$.At $x = A / 2$,the potential energy is $U = \frac{1}{2} k (A / 2)^2 = \frac{1}{4} (\frac{1}{2} k A^2) = \frac{1}{4} 	imes 25 \ J = 6.25 \ J$.The kinetic energy $K$ at any position is given by $K = E - U$.Substituting the values,$K = 25 \ J - 6.25 \ J = 18.75 \ J$.

The maximum potential energy of a block executing simple harmonic motion is $25 \ J$. $A$ is the amplitude of oscillation. At $x = A / 2$,the kinetic energy of the block is $...............$ (in $J$)

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