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(B) The restoring force in $SHM$ is given by $F = -kx$,where $k$ is the force constant and $x$ is the displacement from the mean position.At the extre

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$2$

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(B) The restoring force in $SHM$ is given by $F = -kx$,where $k$ is the force constant and $x$ is the displacement from the mean position.At the extreme position,the displacement $x = A = 2 \, cm = 0.02 \, m$.The force at the extreme position is $F_{ext} = kA = 4 \, N$.Therefore,$k = \frac{4}{0.02} = 200 \, N/m$.At the midway point between the mean position $(x = 0)$ and the extreme point $(x = A)$,the displacement is $x = \frac{A}{2} = \frac{2 \, cm}{2} = 1 \, cm = 0.01 \, m$.The force at this point is $F = kx = 200 	imes 0.01 = 2 \, N$.

The amplitude of a particle executing $SHM$ is $2 \, cm$ and the force acting at the extreme position on the particle is $4 \, N$. What is the force at the midway point between the mean position and the extreme point (in $, N$)?

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