Question

Calculate the time period and oscillation frequency of a
simple pendulum of length 20 cm at a certain place, where
g = 9.8 m s-2.​

Answer 1

A simple pendulum consists of a Bob attached with a string to a fixed support and the Bob is slightly displaced .

The time period of a pendulum refers to the time taken by the oscillatory body (that is the bob) to complete one full oscillation.

The frequency of the pendulum refers to the rapidity with which the oscillatory body completes its oscillations. It is the reciprocal of time period

$T = 2\pi \sqrt{\dfrac{l}{g} }$

$= > T = 2\pi \sqrt{\dfrac{0.2}{9.8} }$

$= > T = 2\pi \sqrt{\dfrac{1}{49} }$

$= > T = 2\pi \times \dfrac{1}{7}$

$= > T = \dfrac{2\pi}{7} \: sec$

Now frequency will be reciprocal of Time period :

Freq. = $\dfrac{1}{T} = \dfrac{7}{2\pi}$ Hz