Question

Two pendulums of lengths $1m$ and $1.21m$ respectively start swinging together with same amplitude. The number of vibrations that will be executed by the longer pendulum before the two will swing together again are

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

Answer: (B)

Let $T_1$ and $T_2$ be the time periods of the pendulum with lengths $1.0m$ and $1.21m$ respectively
$\frac{T_2}{T_1}=\sqrt{\frac{l_2}{l_1}}=\sqrt{\frac{1.21}{1}}=1.1$ ......(i)
Let $v$ and $v$ be the vibrations made by two pendulum to swing together.
$\therefore v_1{T}_1=v_2{T}_2$ .......(ii)
For the two pendulum to swing together, required condition is
$v_1-v_2=1$
or $v_1=v_2+1$
$\therefore \left(v_2+1\right)T_1=v_2{T}_2$
or $\left(v_2+1\right)v_2=T_2/T_1=1.1$
or $1+\frac{1}{v_2}=1.1$
or $\frac{1}{v_2}=1.1-0.1=\frac{1}{10}$
or $v_2=10$