Answer to Question #262075 in Physics for lowrayne

Question

A simple pendulum is found to vibrate 50 times within 200s. When 1.5 m of its

length is reduced to certain length, it vibrates 50 times in 175 s. Find the

original length of the pendulum.

Answer 1

The period of the original pendulum:

$$T_1/N=2\pi\sqrt{L/g},\\\space\\ T_2/N=2\pi\sqrt{(L-l)/g}.\\\space\\ \frac{T_1}{T_2}=\sqrt{\frac L{L-l}},\\\space\\ \bigg(\frac{T_1}{T_2}\bigg)^2=\frac L{L-l},\\\space\\ L=l·\frac{T_1^2/T_2^2}{T_1^2/T_2^2-1}=6.4\text{ m}.$$