Question

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

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

Answer 1

The correct option is D $\frac{21}{8}\text{s}$

Equation of displacement of pendulums performing $\text{SHM}$ and starting from extreme position is,

pendulum 1: $y_1=A\sin ({\omega }_{1}t+\frac{\pi }{2})...\left(1\right)$
Equation of motion of pendulum 2: $y_2=A\sin ({\omega }_{2}t-\frac{\pi }{2})...\left(2\right)$
Since on putting $t=0$ we get $y_1=A,\&y_2=-A$

For the pendulums to vibrate in same phase :
$\Delta \varphi =0$
$\Rightarrow {\varphi }_1={\varphi }_2$
${\omega }_{1}t+\frac{\pi }{2}={\omega }_{2}t-\frac{\pi }{2}$
$\Rightarrow t=\frac{\pi }{{\omega }_2-{\omega }_1}$
Substituting $\omega =\frac{2\pi }{T}$
$\Rightarrow t=\frac{T_1{T}_2}{2(T_1-T_2)}=\frac{7\times 3}{2(7-3)}$
$\therefore t=\frac{21}{8}\text{s}$