Question 20: Oscillations

Question

A uniform disc of radius $R$ and mass $M$ is free to oscillate about the axis $A$ as shown in the figure. For small oscillations the time period is $\_\_\_\_$ .
(g is acceleration due to gravity)

Accepted Answer

Given,
$$ x = a \sin \left(50t + \frac{\pi}{3}ight) $$
This is a case of simple harmonic motion, where
$$ \omega = 50 	ext{ rad/s} $$
We need to find:
$t_1$: time when particle comes to rest, i.e. velocity $v=0$
$t_2$: time when acceleration is zero, i.e. $a=0$
First, velocity:
$$ v = \frac{dx}{dt} = a \cdot 50 \cos\left(50t + \frac{\pi}{3}ight) $$
For the particle to come to rest,
$$ v=0 $$
So,
$$ \cos\left(50t + \frac{\pi}{3}ight)=0 $$
Now, $\cos 	heta = 0$ when
$$ 	heta = \frac{\pi}{2}, \frac{3\pi}{2}, \dots $$
Taking the first time,
$$ 50t + \frac{\pi}{3} = \frac{\pi}{2} $$
$$ 50t = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} $$
$$ t_1 = \frac{\pi}{300}	ext{ s} $$
Now, acceleration in SHM is
$$ \alpha = \frac{d^2x}{dt^2} = -\omega^2 x $$
So acceleration becomes zero when displacement becomes zero:
$$ x=0 $$
That means,
$$ a\sin\left(50t + \frac{\pi}{3}ight)=0 $$
So,
$$ \sin\left(50t + \frac{\pi}{3}ight)=0 $$
Now, $\sin	heta=0$ when
$$ 	heta = 0, \pi, 2\pi, \dots $$
We take the first positive time. So,
$$ 50t + \frac{\pi}{3} = \pi $$
$$ 50t = \pi - \frac{\pi}{3} = \frac{2\pi}{3} $$
$$ t_2 = \frac{2\pi}{150} = \frac{\pi}{75}	ext{ s} $$
Hence,
$$ t_1 = \frac{\pi}{300}	ext{ s}, \quad t_2 = \frac{\pi}{75}	ext{ s} $$
So the correct option is:
$$ \boxed{	ext{Option A}} $$

Answer

$$ 2 \pi \sqrt{\frac{5 R}{4 g}} $$

Answer

$$ 2 \pi \sqrt{\frac{2 R}{3 g}} $$

Answer

$$ 2 \pi \sqrt{\frac{3 R}{2 g}} $$

Answer

$$ 2 \pi \sqrt{\frac{3 R}{g}} $$

Oscillations

A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is \_\_\_\_ . (g is acceleration due to gravity)

Choose the correct answer: