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 s⁻².

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Solution

It is given that:
Time period of the second pendulum, T = 2 s
Acceleration due to gravity of a given place, g = $π$ ² ms⁻²
The relation between time period and acceleration due to gravity is given by,
$T = 2 π \sqrt{(\frac{l}{g})}$
where l is the length of the second pendulum.

Substituting the values of T and g, we get:
$\Rightarrow 2 = 2 π \sqrt{(\frac{l}{π^{2}})} \Rightarrow \frac{1}{π} = \frac{\sqrt{l}}{π} \Rightarrow l = 1 m$

Hence, the length of the pendulum is 1 m.

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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 s−2.

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 s⁻².