Question

A simple pendulum is found to vibrate 50 times within 200s. When 1.5 meter of its length is reduced to a certain length it vibrates 50 times in 175s. Find the original length of the pendulum.

Answer 1

Given:

$N_1=50,\quad t_1=200\:\rm s$

$N_2=50,\quad t_1=175\:\rm s$

The period of oscillation of a simple pendulum

$$T_1=\frac{t_1}{N_1}=2\pi\sqrt{l/g}$$ $$T_2=\frac{t_2}{N_2}=2\pi\sqrt{(l-x)/g}$$

$$\left(\frac{t_1}{t_2}\right)^2=l/(l-x)$$ $$\left(\frac{200}{175}\right)^2=l/(l-1.5)$$ $$16/9=l/(l-1.5)$$

$$16l-24=9l$$

$$l=24/7=3.4\:\rm m$$