Question

Two pendulums of time periods $3s$ and $7s$ respectively start oscillating simultaneously from two opposite extreme positions. After how much time they will be in phase

• A. $\frac{21}{8}s$
• B. $\frac{21}{4}s$
• C. $\frac{21}{2}s$
• D. $\frac{21}{10}s$

Answer 1

Answer: (A)

$\frac{21}{8}s$

Given $T_1=3$ s and $T_2=7$ s

Thus, ${\omega }_1=\frac{2\pi }{3}$ rad/s and ${\omega }_2=\frac{2\pi }{7}$ rad/s

Also, the initial phase difference is ${\varphi }_{2,0}-{\varphi }_{1,0}=\pi$

The phase of the first pendulum is ${\varphi }_1={\varphi }_{1,0}+{\omega }_{1}t$

and, phase of second pendulum is ${\varphi }_2={\varphi }_{2,0}+{\omega }_{2}t$

Let after a time $t$, both the pendulums are in phase.

Thus, ${\varphi }_2-{\varphi }_1=0\Rightarrow ({\omega }_2-{\omega }_1)t+{\varphi }_{2,0}-{\varphi }_{1,0}=0$

Thus, $(\frac{2\pi }{3}-\frac{2\pi }{7})t=\pi \Rightarrow t=\frac{21}{8}\text{s}$