JEE Mains · Physics · STD 11 - 13. oscillations
In figure \((A),\) mass ' \(2 m\) ' is fixed on mass ' \(m\) ' which is attached to two springs of spring constant \(k\). In figure \((B),\) mass ' \(m\) ' is attached to two spring of spring constant ' \(k\) ' and ' \(2 k\) '. If mass ' \(m\) ' in \((A)\) and \((B)\) are displaced by distance ' \(x\) ' horizontally and then released, then time period \(T_{1}\) and \(T_{2}\) corresponding to \((A)\) and \((B)\) respectively follow the relation.

- A \(\frac{T_{1}}{T_{2}}=\frac{3}{\sqrt{2}}\)
- B \(\frac{ T _{1}}{ T _{2}}=\sqrt{\frac{3}{2}}\)
- C \(\frac{ T _{1}}{ T _{2}}=\sqrt{\frac{2}{3}}\)
- D \(\frac{ T _{1}}{ T _{2}}=\frac{\sqrt{2}}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{T_{1}}{T_{2}}=\frac{3}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(T _{1}=2 \pi \sqrt{\frac{3 m }{2 k }}\) \(T _{2}=2 \pi \sqrt{\frac{ m }{3 k }}\) \(\frac{ T _{1}}{ T _{2}}=\frac{2 \pi \sqrt{\frac{3 m }{2 k }}}{2 \pi \sqrt{\frac{ m }{3 k }}}=\frac{3}{\sqrt{2}}\)
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