Question
Two pendulums of lengths 100cm and 110.25cm start oscillating in phase simultaneously. After how many oscillations will they again be in phase together?
✓
Answer
$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$$\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.
$\text{l}_2=110.25\text{cm}$
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.
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