Question

A simple pendulum is of length 50cm. Find its time period and frequency of oscillation.(g $9.8m/s^2$ )

• A. $2.419\text{sec,}0.8045Hz$
• B. $3.219\text{sec,}0.5045Hz$
• C. $5.619\text{sec,}0.6045Hz$
• D. $1.419\text{sec,}0.7045Hz$

Answer 1

The correct option is D $1.419\text{sec,}0.7045Hz$

Length of pendulum = 50 cm

$\frac{50}{100}m=\frac{1}{2}m$

Given $g=9.8m/s^2$

Time period of simple pendulum

$\begin{array}{c}T=2\pi \sqrt{\frac{\ell }{g}}\\ =2\pi \sqrt{\frac{Y2}{9.8}}\\ =2\pi \sqrt{\frac{1}{2\times 9.8}}\sec \\ =2\pi \times 0.22\sec \\ =2\times 3.14\times 0.22\\ =1.419\sec \\ Frequencyn=\frac{1}{T}=\frac{1}{1.419}=0.7045Hz\end{array}$

Option D.