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° to the right with the vertical, the other pendulum makes an angle of 1° 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

Answer: (B)

$\frac { 2 } { 3 } \pi$

Answer 2

Let $\theta _ { 1 }$ and $\theta _ { 2 }$ be the angular displacement of first and second pendulums respectively at an instant$\phi _ { 1 }$ and $\phi _ { 2 }$ be the initial phases of the two pendulums.

Then

$\theta _ { 1 } = \theta _ { 0 } \sin \left( \omega t + \phi _ { 1 } \right)$ …… (i)

And $\theta _ { 2 } = \theta _ { 0 } \sin \left( \omega t + \phi _ { 2 } \right)$ …… (ii)

For first pendulum d

From (i) we get

$2 = 2 \sin \left( \omega t + \phi _ { 1 } \right)$ or $\sin \left( \omega t + \phi _ { 1 } \right) = 1$

Or $\omega \mathrm { t } + \phi _ { 1 } = 90 ^ { \circ }$ ….. (iii)

For second pendulum

From (ii), we get

$- 1 = 2 \sin \left( \omega \mathrm { t } + \phi _ { 2 } \right)$ or $\sin \left( \omega \mathrm { t } + \phi _ { 2 } \right) = - \frac { 1 } { 2 }$

Or $\sin \left( \omega t + \phi _ { 2 } \right) = \sin \left( 180 ^ { \circ } + 30 ^ { \circ } \right) = \sin 210 ^ { \circ }$

$\therefore \omega \mathrm { t } + \phi _ { 2 } = 210 ^ { \circ }$ …… (iv)

$\therefore$ subtracting (iii) from (iv), we get

$\left( \omega \mathrm { t } + \phi _ { 2 } \right) - \left( \omega \mathrm { t } + \phi _ { 1 } \right) = 210 ^ { \circ } - 90 ^ { \circ } = 120 ^ { \circ } = \frac { 2 \pi } { 3 }$