Question

A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum, i.e., a pendulum having frequency same as that of the block.

Answer 1

An equivalent simple pendulum has same time period as that of the spring mass system.
The time period of a simple pendulum is given by,

\[T_p = 2\pi\sqrt{\left( \frac{l}{g} \right)}\]

where l is the length of the pendulum, and
           g is acceleration due to gravity.

Time period of the spring is given by,

\[T_s = 2\pi\sqrt{\left( \frac{m}{k} \right)}\]

where m is the mass, and 
           k is the spring constant.

Let x be the extension of the spring.
For small frequency, T_P ​can be taken as equal to T_S.    

\[\Rightarrow \sqrt{\left( \frac{l}{g} \right)} = \sqrt{\left( \frac{m}{k} \right)}\]

\[ \Rightarrow \left( \frac{l}{g} \right) = \left( \frac{m}{k} \right)\]

\[ \Rightarrow l = \frac{mg}{k} = \frac{F}{k} = x\]

(\[\because\] restoring force = weight = mg) 

\[\therefore\] l = x (proved)