JEE Mains · Physics · STD 11 - 13. oscillations
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
- A \(\frac{1}{2 \pi} \sqrt{\frac{ K _1+ K _2}{ m }}\)
- B \(\frac{1}{2 \pi} \sqrt{\frac{ K _1- K _2}{ m }}\)
- C \(2 \pi \sqrt{\frac{ m }{ K _1+ K _2}}\)
- D \(2 \pi \sqrt{\frac{m}{K_1-K_2}}\)
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 \(x\) \(F =-\left( k _1 x+ k _2 x \right)=-\left( k _1+ k _2\right) x\) \(a =\frac{ F }{ m }=-\left(\frac{ k _1+ k _2}{ m }\right) x =-\omega^2 x\)…
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