Question

Two pendulums of lengths $1 \,m$ and $1.21 \,m$ 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 $1.0 \,m$ and $1.21 \, m$ 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 $