Question

A simple pendulum is moving simple harmonically with a period of 6 s between two extreme position B and C about a point O. If the angular distance between B and C is 10 cm, how long will the pendulum take to move from position C to a position D exactly midway between O and C

• A. 0.5 s
• B. 1.0 s
• C. 1.5 s
• D. 3 s

Answer 1

The correct option is B 1.0 s

Given time period T = 6 s
Amplitude $=OC=OB=\frac{1}{2}BC=\frac{10}{2}=5cm$
$\therefore$ OD=2.5 cm
Let the displacement of the pendulum be given by
X=A sin $(\omega +\varphi )$
Where A=5 cm and $\omega =\frac{2\pi }{T}=\frac{2\pi }{6}=\frac{\pi }{3}rad{s}^{-1}$
Let us suppose that at t = 0, the pendulum is at C, i.e., at t = 0, x = A,
So that
A=A sin $(\omega \times 0+\varphi )$
or, A=$Asin\varphi orsin\varphi =1$ or $\varphi =\frac{\pi }{2}$
Thus the motion of the pendulum is given by(putting $\varphi =\frac{\pi }{2}$
$X=Asin(\omega t+\frac{\pi }{2})=Acos\omega t=5cos\omega t$
The value of t for which x = 2.5 cm is given by
2.5 = 5 cos $\omega t$
or cos $\omega$ t =$\frac{1}{2}$ or $\omega t$ =$\frac{\pi }{3}$
Since $\omega =\frac{\pi }{3}$ we have t = 1 second