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A block of mass m, attached to a spring of spring constant k, oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. The block has a speed v when the spring is at its natural length. Before coming to an instantaneous rest, if the block moves a distance x from the mean position, then

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$k = v \sqrt{\frac{m}{k}}$

$x = \frac{1}{v} \sqrt{\frac{m}{k}}$

$x = \sqrt{\frac{mv}{k}}$

$x = \sqrt{\frac{m}{k}}$

answer is C.

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Detailed Solution

$\frac{1}{2} {mv}^{2} = \frac{1}{2} {kx}^{2}$

$∴ x = v \sqrt{\frac{m}{k}}$

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