JEE Mains · Physics · STD 11 - 13. oscillations
In the reported figure, two bodies \(A\) and \(B\) of masses \(200\, {g}\) and \(800\, {g}\) are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be \(.....\,{rad} / {s}\) when \({k}=20 \,{N} / {m} .\)

- A \(100\)
- B \(20\)
- C \(10\)
- D \(30\)
Answer & Solution
Correct Answer
(C) \(10\)
Step-by-step Solution
Detailed explanation
\(\omega=\sqrt{\frac{k_{\text {eq }}}{\mu}}\) \(\mu=\) reduced mass springs are in series connection \(k _{eq}=\frac{ k _{1} k _{2}}{ k _{1}+ k _{2}}\) \(k _{ eq }=\frac{ k \times 4 k }{5 k }=\frac{4 k }{5}\) \(k _{ eq }=\frac{4 \times 20}{5} N / m =16 N / m\)…
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