# Thread: [SOLVED] Time period of simple pendulum(with a catch!)?

1. ## [SOLVED] Time period of simple pendulum(with a catch!)?

Find the time period of a simple pendulum if the mass of the bob is $\displaystyle m$ and mass of the rod is $\displaystyle M$.

Spoiler:
$\displaystyle T = 2\pi\sqrt{\left(\frac{m + \frac{M}{3}}{m + \frac{M}{2}}\left(\frac{L}{g}\right)\right)}$

Anyone?

2. ## Compound Pendulum

Hello fardeen_gen
Originally Posted by fardeen_gen
Find the time period of a simple pendulum if the mass of the bob is $\displaystyle m$ and mass of the rod is $\displaystyle M$.

Spoiler:
$\displaystyle T = 2\pi\sqrt{\left(\frac{m + \frac{M}{3}}{m + \frac{M}{2}}\frac{L}{g}\right)}$

Anyone?
You need the formula for the period of oscillation of a compound pendulum, which is:

$\displaystyle T = 2\pi\sqrt{\frac{I}{mgh}}$

where $\displaystyle I$ is the moment of inertia about the axis of rotation
$\displaystyle m$ is the mass
$\displaystyle h$ is the distance of the centre of mass from the axis of rotation

The moment of inertia of a rod of mass $\displaystyle M$, length $\displaystyle L$ about one end is $\displaystyle \tfrac13ML^2$, and the moment of inertia of the bob is $\displaystyle mL^2$, so $\displaystyle I= \tfrac13ML^2+mL^2$

Total mass = $\displaystyle M+m$

and, taking moments about the axis of rotation:

$\displaystyle (m+M)h = \tfrac12ML+mL$

$\displaystyle \Rightarrow T =2\pi\sqrt{\frac{\tfrac13ML^2+mL^2}{g(\tfrac12ML+m L)}}$

$\displaystyle =2\pi\sqrt{\frac{(\tfrac13M+m)L}{g(\tfrac12M+m)}}$