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Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :
Option: 1
$9$

Option: 2
$10$

Option: 3
$8$

Option: 4
$11$

Answers (1)

best_answer

$\mathrm{T_{1}=2 \pi \sqrt{\frac{l_{1}}{g}}=\frac{2 \pi}{\omega_{1}}=\frac{2 \pi \times \sqrt{1 \cdot 21}}{\sqrt{g}}} \\$

$\mathrm{T_{2}=2 \pi \sqrt{\frac{l_{2}}{g}}=\frac{2 \pi}{\omega_{2 }}=\frac{2 \pi}{\sqrt{g}}(\sqrt{1})}$

Let $\theta_{1}$ and $\theta_{2}$ be the phase covered by the 1st and 2nd pendulum respectively. The two pendulums to be in the same phase,

$\theta_{2}-\theta_{1}=2 n \pi$

$(n=1,2,3......)$

For the minimum number of vibrations ie. $(n=1)$

$\mathrm{\theta_{2}-\theta_{1}=\omega_{2} t-\omega_{1} t=2 \pi} \\$

$\mathrm{t\left(\frac{2 \pi}{T_{2}}-\frac{2 \pi}{T_{1}}\right)=2 \pi}$

$\mathrm{\frac{1}{t} =\frac{1}{T_{2}}-\frac{1}{T_{1}}} \\$

$\because \mathrm{\frac{T_{1}}{T_{2}} =\frac{1.1}{1}=\frac{11}{10}} \\$

$\mathrm{\therefore T_{1} =\frac{11 T_{2}}{10}}$

$\mathrm{\frac{1}{t} =\frac{1}{T_{2}}-\frac{1}{\frac{11 T_{2}}{10}}} \\$

$\mathrm{\frac{1}{t} =\frac{1}{11 T_{2}}} \\$

$\mathrm{t=11 T_{2} }$

Hence the correct option is 4

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Riya

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