Question

A simple pendulum of length $40cm$ oscillates with an angular amplitude of $0.04rad$. Find the angular acceleration when the bob is in momentary rest. Take $g=10m/s^2$.

• A. $2rad/s^2$
• B. $3rad/s^2$
• C. $1rad/s^2$
• D. $4rad/s^2$

Answer 1

Answer: (C)

$1rad/s^2$

$\text{Given:-}$ Length of simple pendulum = 40cm

Angular amplitude = 0.04 rad

$\text{To find:-}$ The angular acceleration.

$\text{Solution:-}$

The angular frequency is

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

$\omega =\sqrt{\frac{10m/s^2}{0.4m}}$

$\omega =5se{c}^{-1}$

the time period is

$\frac{2\pi }{\omega }=\frac{2\pi }{5se{c}^{-1}}=1.26sec.$

Linear amplitude = $40cm\times 0.04=1.6cm$

Angular speed at displacement 0.02 rad is

$\omega =\left(5se{c}^{-1}\right)\sqrt{\left({0.04}^2\right)-\left({0.02}^2\right)}rad=0.17rad/sec$

where speed of the bob at this instant

$=\left(40cm\right)\times {\left(0.175\right)}^{-1}=6.8cm/sec$

At momentary rest, the bob is in extreme position.

Thus, the angular acceleration

$\alpha =\left(0.4rad\right)\left(25se{c}^2\right)=1rad/se{c}^2$

$\text{Hence the correct option is C}$