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.

#### Solution

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)