JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2023
- Medium
- MCQ
A mass \(m\) is attached to two springs as shown in figure. The spring constants of two springs are \(K _1\) and \(K _2\). For the frictionless surface, the time period of oscillation of mass \(m\) is

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(2 \pi \sqrt{\frac{ m }{ K _1+ K _2}}\)
Step-by-step Solution
Detailed explanation
On displacing \(m\) to right by \(x\)
\(F =-\left( k _1 x+ k _2 x \right)=-\left( k _1+ k _2\right) x\)
\(a =\frac{ F }{ m }=-\left(\frac{ k _1+ k _2}{ m }\right) x =-\omega^2 x\)
\(\therefore \omega=\sqrt{\frac{ k _1+ k _2}{ m }} \Rightarrow T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{ m }{ k _1+ k _2}}\)
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