Question

the time period of a small oscillation of a compound pendulum varies with dist. x between the point of suspension and the center of gravity​

Answer 1

Answer:

1

We know

$\tau = F.r$

Now we calculate torque about AOR , so

$\tau = - Fx \sin( \theta) = I \alpha \\ \\ for \: small \: \theta \implies\: sin \theta \approx \theta \\ so \\ \alpha = - \frac{ Fx}{I } \theta$

Since it's shm so

$\alpha = - { \omega}^{2} \theta = - \frac{ Fx}{I } \theta \\ \\ \omega = \sqrt{\frac{ Fx}{I } } \\ \\ t = \frac{2 \pi}{ \omega} = \frac{2\pi}{ \sqrt{\frac{ Fx}{I } } }$

So

$\red{ \boxed{\red {\huge{t \: \: \: \: \alpha \: \: \: \: \frac{1}{ \sqrt{x} } }}}}$

Option first is right