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  • Two pendulum of length 256 cm and 169 cm start vibrating in phase. At some instant, the two are at their mean position in the sane phase. The minimum number of vibrations of the shorter pendulum af

Two pendulum of length 256 cm and 169 cm start vibrating in phase. At some instant, the two are at their mean position in the sane phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position?

Option: 1

18


Option: 2

14


Option: 3

15


Option: 4

16


Answers (1)

best_answer

\begin{aligned} & L_1=256 \mathrm{~cm}=\frac{256}{100}=2.56 \mathrm{~m} \\ & L_2=169 \mathrm{~cm}=\frac{169}{100}=1.69 \mathrm{~m} \end{aligned}

\begin{aligned} & T_1=\text { longer Pendulum } \\ & T_2=\text { Shorter Pendulum } \\ & T=2 \pi \sqrt{\frac{l}{g}} \\ & \frac{T_1}{T_2} \propto \sqrt{\frac{L_{1}}{L_2}} \\ & \frac{T_1}{T_2} \propto \sqrt{\frac{2.56}{1.69}}=\frac{16}{13} \\ & \frac{T_1}{T_2}=\frac{16}{13} \\ & 10 T_1=16 T_2 \end{aligned}

Vibration of shorter Pendulum is = 16

Posted by

Pankaj Sanodiya

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