Answer to Question #293136 in Physics for joy

Question

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

Answer 1

Given:

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

$N_2=50,\; t_2=175\: \rm s$

$l_1=l$

$l_2=l-1.5$

The period of oscillation of a simple pendulum is given by

$$T=\frac{t}{N}=2\pi\sqrt{l/g}$$

Hence

$$\frac{t_1}{N_1}=2\pi\sqrt{l/g}$$

$$\frac{t_2}{N_2}=2\pi\sqrt{(l-1.5)/g}$$

We get

$$(t_1/t_2)^2=l/(l-1.5)$$

$$64/49=l/(l-1.5)$$

$$l=6.4\:\rm m$$