JEE Mains · Physics · STD 11 - 13. oscillations
Consider two identical springs each of spring constant \(k\) and negligible mass compared to the mass \(M\) as shown. Fig. \(1\) shows one of them and Fig. \(2\) shows their series combination. The ratios of time period of oscillation of the two \(SHM\) is \(\frac{ T _{ b }}{ T _{ a }}=\sqrt{ x },\) where value of \(x\) is (Round off to the Nearest Integer)

- A \(3\)
- B \(2\)
- C \(6\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(T _{ a }=2 \pi \sqrt{\frac{ M }{ K }}\) \(T _{ b }=2 \pi \sqrt{\frac{ M }{ K / 2}}\) \(\frac{ T _{ b }}{ T _{ a }}=\sqrt{2}=\sqrt{ x }\) \(\Rightarrow x =2\)
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