Question

Two simple pendulums of lengths $1m$ and $16m$ respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed $n$ oscillations where $n$ is

• A. $1\frac{1}{3}$
  
• B. $\frac{1}{4}$
  
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

lets say they are again in phase when shorter pendulum has made $n_1$ oscillations and bigger pendulum $n_2$ oscillations. so total time elapsed $=T_1\times n_1=T_2\times n_2$

$\Rightarrow \frac{n_1}{n_2}=\frac{T_2}{T_1}=\sqrt{\frac{l_2}{l_1}}=\sqrt{\frac{16}{1}}=\frac{4}{1}$

they will again be in phase for the first time when the shorter pendulum has made 4 oscillations and the longer pendulum has made 1 oscillation.