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(D) The total energy $E$ in $S.H.M.$ is given by $E = \frac{1}{2} k A^2$, where $k = m \omega^2$ is the force constant and $A$ is the amplitude.Given

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$25 \sqrt{2} \,N$

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(D) The total energy $E$ in $S.H.M.$ is given by $E = \frac{1}{2} k A^2$, where $k = m \omega^2$ is the force constant and $A$ is the amplitude.Given that the kinetic energy $K$ and potential energy $U$ are equal, $K = U = \frac{E}{2}$.The potential energy at displacement $x$ is $U = \frac{1}{2} k x^2$.Thus, $\frac{1}{2} k x^2 = \frac{1}{2} (\frac{1}{2} k A^2) \implies x^2 = \frac{A^2}{2} \implies x = \frac{A}{\sqrt{2}}$.The force acting on the particle at displacement $x$ is $F = -kx = -m \omega^2 x$.The maximum force is $F_m = k A = 50 \,N$.The magnitude of the force at displacement $x$ is $F' = kx = k (\frac{A}{\sqrt{2}}) = \frac{F_m}{\sqrt{2}}$.Substituting $F_m = 50 \,N$, we get $F' = \frac{50}{\sqrt{2}} = 25 \sqrt{2} \,N$.

$A$ body is executing $S.H.M.$ under the action of a force having a maximum magnitude of $50 \,N$. When its energy is half kinetic and half potential, the magnitude of the force acting on the particle is:

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