Question

A small iron ball of mass m is suspended with the help of mass rod of length l is free to oscillate in vertical plane then its time period of oscillation is

Answer 1

Answer:

Time period of the motion of pendulum is given as

$T = 2\pi\sqrt{\frac{L}{g}}$

Explanation:

As we know that mass m is suspended by a rod of length L

so here we have

$\tau = I\alpha$

$\tau = mg sin\theta L$

now for small angle we have

$mgL\theta = I \alpha$

so we have

$\alpha = \frac{I}{mgL} \theta$

$\alpha = \frac{mL^2}{mgL}\theta$

$\alpha = \frac{L}{g} \theta$

so we have

$\omega = \sqrt{\frac{L}{g}}$

so the time period of the motion is

$T = 2\pi\sqrt{\frac{L}{g}}$

#Learn

Topic : SHM of pendulum

https://brainly.in/question/5108862