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Question
The spring shown in figure is unstretched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F, find (a) the amplitude and the time period of the motion of the block, (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position.
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Solution
(a) We know-
f = kx
\[\Rightarrow x = \frac{F}{k}\]
\[\text { Acceleration }= \frac{F}{m}\]
Using the relation of time period of S.H.M.,
\[\text { Time period }, T = 2\pi\sqrt{\frac{\text { Displacement } }{\text { Acceleration }}}\]
\[ = 2\pi\sqrt{\frac{\left( \frac{F}{k} \right)}{\left( \frac{F}{m} \right)}} = 2\pi\sqrt{\frac{m}{k}}\]
Amplitude = Maximum displacement \[= \frac{F}{k}\]
When the block passes through the equilibrium position, the energy contained by the spring is given by,
\[E = \frac{1}{2}k x^2 = \frac{1}{2}k \left( \frac{F}{k} \right)^2 = \frac{1}{2}\left( \frac{F^2}{k} \right)\]
(b) At the mean position, potential energy is zero.
Kinetic energy is given by,
\[\frac{1}{2}k x^2 = \frac{1}{2}\frac{F^2}{k}\]
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