Question

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

Answer 1

Let's first find the period of vibrations of first and second pendulum:

$$T_1=\dfrac{t_1}{N}=\dfrac{200\ s}{50}=4\ s,$$ $$T_2=\dfrac{t_2}{N}=\dfrac{175\ s}{50}=3.5\ s.$$

From the definition of the period of pendulum, we have:

$$T_1=2\pi\sqrt{\dfrac{L}{g}},$$ $$T_2=2\pi\sqrt{\dfrac{L-l}{g}}.$$

Let's divide the first expression by the second one:

$$\dfrac{T_1}{T_2}=\sqrt{\dfrac{Lg}{g(L-l)}},$$ $$\dfrac{4}{3.5}=\sqrt{\dfrac{L}{(L-l)}},$$ $$L=1.31(L-l),$$ $$L=\dfrac{1.31l}{0.31}=\dfrac{1.31\cdot1.5\ m}{0.31}=6.34\ m.$$