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  • In figure <img alt="\mathrm{(A)}" src="/latex-image/?%5Cmat

In figure \mathrm{(A)}, mass ' 2 \mathrm{~m}^{\prime} is fixed on mass ' \mathrm{m} ' which is attached to two springs of spring constant \mathrm{k}. In figure \mathrm{(B)}, mass ' \mathrm{m} ' is attached to two springs of spring constant ' \mathrm{k} ' and ' \mathrm{ \mathrm{2k}} '. If mass ' \mathrm{m}$ ' in $(\mathrm{A})$ and in $(\mathrm{B}) are displaced by distance \mathrm{' x\: '}horizontally and then released, then time period \mathrm{T_{1} \: and\: T_{2}} corresponding to \mathrm{(A) and (B)} respectively follow the relation.

Option: 1

\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{3}{\sqrt{2}}


Option: 2

\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{3}{2}}
 


Option: 3


\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{2}{3}}

 


Option: 4

\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{\sqrt{2}}{3}


Answers (1)

best_answer

Equivalent of figure (A) is

\mathrm{k_{1}=k+k=2k}

Equivalent of figure (B) is

\mathrm{k_{2}=k+2k=3k}

\mathrm{T_1=2 \pi \sqrt{\frac{3 m}{k_1}}=2 \pi \sqrt{\frac{3 m}{2 k}} }

\mathrm{T_2=2 \pi \sqrt{\frac{m}{k_2}}=2 \pi \sqrt{\frac{m}{3 k}} }

\mathrm{\frac{T_1}{T_2}=\frac{3}{\sqrt{2}}}

Hence (1) is correct option.
 

Posted by

avinash.dongre

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