Oscillations — Physics STD 11 Science — Question

$\text{l}_1=100\text{cm}$

$\text{l}_2=110.25\text{cm}$

For smaller pendulum, $\text{T}_1=2\pi\sqrt{\frac{100}{\text{g}}}\cdots\text{(i)}$

For larger pendulum, $\text{T}_2=2\pi\sqrt{\frac{110.25}{\text{g}}}\cdots\text{(ii)}$

Let these pendulums oscillate in phase again if larger pendulum completes 'n' oscillations. It means smaller pendulum must complete (n + 1) oscillations.

$\text{nT}_2=(\text{n}+1)\text{T}_1$

$\frac{\text{n}+1}{\text{n}}$

$\frac{\text{T}_2}{\text{T}_1}=\sqrt{\frac{110.25}{100}}$

$=1.05$

$=1+\frac{1}{\text{n}}$

$=1.05$

$=\frac{1}{\text{n}}=0.05$

$=\frac{5}{100}=\frac{1}{20}$

$\therefore\text{n}=20$

Hence both pendulums will again oscillate in phase after 20 oscillations of the larger or 21 oscillations of the smaller pendulum.