Question

A simple pendulum has a period of 4.2 secs. When the length of the pendulum is shortened by 1 meter, the period is 3.7 sec. Calculate the original length. The value of acceleration due to gravity.

Answer 1

$$T_1=2\pi \sqrt{\frac{l}{g}}$$

$$T_2=2\pi \sqrt{\frac{l-x}{g}}$$

$$\frac{l-x}{l}=(\frac{T_2}{T_1})^2$$

$$\frac{l-1}{l}=(\frac{3.7}{4.2})^2$$ $$l = 4.5 m.$$ $$g=l(\frac{2 \pi}{T_1})^2=4.5(\frac{2 \pi}{4.2})^2=10 m/s^2$$