Question

A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a second pendulum at a place where g = π² m/s².

Answer 1

It is given that:
Time period of the second pendulum, T = 2 s
Acceleration due to gravity of a given place, g =\[\pi^2\]ms⁻²

^{The relation between time period and acceleration due to gravity is given by,}

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

where l is the length of the second pendulum.

Substituting the values of T and g, we get:

\[\Rightarrow 2 = 2\pi\sqrt{\left( \frac{l}{\pi^2} \right)}\] 

\[ \Rightarrow \frac{1}{\pi} = \frac{\sqrt{l}}{\pi}\] 

\[ \Rightarrow l = 1  m\]

Hence, the length of the pendulum is 1 m.