Question

Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^o$ to the right of the vertical, the other pendulum makes an angle of $1^o$ to the left of the vertical. The phase difference between the pendulums is:

• A. $\frac{\pi }{2}$
• B. $\frac{2}{3}\pi$
• C. $\frac{3}{2}\pi$
• D. $\pi$

Answer 1

The correct option is B $\frac{2}{3}\pi$

For simple harmonic motion,

$x=Asin(\omega t+\delta )$

where

$x=$ displacement

$\omega =$ angular velocity

$A=$ amplitude

$\delta =$ phase

Given:

For both the pendulum the amplitude given is same and therefore they are identical, their length are the same and thus their angular velocity are also same as for the pendulum $\omega =\sqrt{\frac{g}{L}}$

$g$ acceleration due to gravity and length of the pendulum is given as $L$.

$\therefore$ for pendulum $1$ we can write,

$x_1=Asin(\omega t+{\delta }_1)$

At $2^0,x_1=A$

$\therefore sin(\omega t+{\delta }_1)=1$

$⟹\omega t+{\delta }_1={90}^0$ .......(1)

Now, Pendulum $2$ can be written as

$x_2=Asin(\omega t+{\delta }_2)$

At $1^0,x_2=-\frac{A}{2}$

$\therefore sin(\omega t+{\delta }_2)=-\frac{1}{2}$

$⟹\omega t+{\delta }_2=-{30}^0$ .....(2)

To find phase difference we are subtracting equation (2) from equation (1),

That is,

${\delta }_1-{\delta }_2={90}^0-(-{30}^0)={120}^0=\frac{2}{3}\pi$

$\therefore$ The phase difference between the two pendulum is $\frac{2}{3}\pi$