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Question: A block of mass m, attached to a spring of spring constant k, oscillates on a smooth horizontal tabl...

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

A

x=m/kx = \sqrt { m / k }

B

x=1vm/kx = \frac { 1 } { v } \sqrt { m / k }

C

x=vm/kx = v \sqrt { m / k }

D

x=mv/kx = \sqrt { m v / k }

Answer

x=vm/kx = v \sqrt { m / k }

Explanation

Solution

By using conservation of mechanical energy

12kx2=12mv2x=vm/k\frac { 1 } { 2 } k x ^ { 2 } = \frac { 1 } { 2 } m v ^ { 2 } \Rightarrow x = v \sqrt { m / k }