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  • In figure (A) mass  is fined on mass ' <img alt="2 \mathrm{~m}" src="https://entran

In figure (A) mass 4 \mathrm{~m} is fined on mass ' 2 \mathrm{~m} ' which is attached to two spring of spring constant 4K. In figure (B) mass ' m ' is attached to two spring of spring constant ' 3K ' and k. If mass 2 m in (A) and in (B) are displaced by distance ' y ' horizontally and then released then time period T_{1}$ and $T_{2} corresponding to (A) and (B) respectively follow the relation 

Option: 1

\frac{T_2}{T_1}=\frac{2}{5}


 


Option: 2

\frac{T_{1}}{T_{2}}=\frac{\sqrt{5}}{1}


Option: 3

\frac{T_{2}}{T_{1}}=\frac{\sqrt{3}}{1}


Option: 4

\frac{T_{1}}{T_{2}}=\frac{\sqrt{3}}{1}


Answers (1)

best_answer

   

\begin{aligned} & T_{1}=2 \pi \sqrt{\frac{m}{k}} \\ & T_1=2 \pi \sqrt{\frac{6 m}{8 K}} \\ & T_{1}=2 \pi \sqrt{\frac{3 m}{4 k}} \\ & T_{1}=\frac{2 \pi \times 1}{2} \sqrt{\frac{3 m}{k}} \\ & T_{1}=\pi \sqrt{\frac{3 m}{k}} \\ \end{aligned}

And,

\begin{aligned} & T_{2}=2 \pi \sqrt{\frac{2 m}{8 K}} \\ & T_{2}=2 \pi \sqrt{\frac{m}{4 k}} \\ & T_{2}=2 \pi \times \frac{1}{2} \sqrt{\frac{m}{K}} \\ & T_{2}=\pi \sqrt{\frac{m}{k}} \\ \end{aligned}so,

\begin{aligned} &\therefore \frac{T_{1}}{T_{2}}=\frac{\frac{\sqrt{3 m}}{\sqrt{k}}}{\frac{\sqrt{m}}{\sqrt{k}}} \\ & =\frac{\sqrt{3 m}}{\sqrt{k}} \times \frac{\sqrt{k}}{\sqrt{m}} \\ & \therefore \frac{T_{1}}{T_{2}}=\frac{\sqrt{3 m}}{\sqrt{m}}=\frac{\sqrt{3}}{1} \text {. } \\ & \frac{T_{1}}{T_{2}}=\frac{\sqrt{3}}{1} \end{aligned}

Posted by

SANGALDEEP SINGH

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