Question

A simple pendulum of length $40cm$ oscillates with an angular amplitude of $0.04rad$. Find the speed of the bob when the string makes $0.02rad$ with the vertical.

• A. $4.2cm/s$
• B. $3.4cm/s$
• C. $6.8cm/s$
• D. $13.6cm/s$

Answer 1

The correct option is C $6.8cm/s$

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

Angular amplitude = 40 rad.

$\text{To find:-}$The speed of the bob

$\text{Solution:-}$

(a)The angular frequency is

$\omega =\sqrt{\frac{g}{l}}=\sqrt{\frac{10m/s^2}{0.4m}}=5s^{-1}$

The time period is $\frac{2\pi }{\omega }=\frac{2\pi }{5s-1}=1.26s$

(b) Linear amplitude$=40cm\times 0.04=1.6cm$

(c) Angular speed $\dot{\theta }=\omega \sqrt{{\theta }_0^2-{\theta }^2}$

$\dot{\theta }=\left(5s^{-1}\right)\sqrt{(0.04{)}^2-(0.02{)}^2}rad=0.17rad/s$

where speed of the bob at this instant

$V=\dot{\theta }A=\left(40cm\right)\times \left(0.17s\right)=6.8cm/s.$

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