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(C) The maximum force on a particle performing simple harmonic motion $(SHM)$ is given by $F_{\max} = m \omega^2 a = 10 \,N$, where $m$ is the mass, $

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$5 \,N$

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(C) The maximum force on a particle performing simple harmonic motion $(SHM)$ is given by $F_{\max} = m \omega^2 a = 10 \,N$, where $m$ is the mass, $\omega$ is the angular frequency, and $a$ is the amplitude.At the mean position, the displacement $y = 0$. At the extreme position, the displacement $y = a$.The force on the particle at any displacement $y$ is given by $F = m \omega^2 y$.When the particle is midway between the mean and extreme positions, the displacement is $y = \frac{a}{2}$.Substituting this value into the force equation:$F = m \omega^2 \left(\frac{a}{2}ight) = \frac{1}{2} (m \omega^2 a)$.Since $m \omega^2 a = 10 \,N$, we get:$F = \frac{1}{2} 	imes 10 \,N = 5 \,N$.

The maximum force acting on a particle executing simple harmonic motion is $10 \,N$. The force on the particle when it is midway between mean and extreme positions will be

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