Question

Two pendulums $X$ and $Y$ of time periods $4s$ and $4.2s$ are made to vibrate simultaneously. They are initially in same phase. After how many vibrations of $X,$ they will be in the same phase again?

• A. $30$
• B. $25$
• C. $21$
• D. $26$

Answer 1

The correct option is C $21$

${\omega }_{1}t={\omega }_{2}t+2\pi$ $({\theta }_1={\theta }_2+2\pi )$
$\therefore t=\frac{2\pi }{{\omega }_1-{\omega }_2}=\frac{2\pi }{\left(2\pi /T_1\right)-\left(2\pi /T_2\right)}$
$t=\frac{T_1{T}_2}{T_2-T_1}$
$t=\frac{4\times 4.2}{4.2-4}=84$
Number of vibration of $X$ in this time are,
$N_1=\frac{t}{T_1}=\frac{84}{4}=21$