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MHT CET 2025 (25 Apr, Shift 2) · Spring Systems

Spring Systems question from MHT CET 2025 (25 Apr, Shift 2)

Q34MHT CET 2025 · 25 APR · SHIFT 2Medium

Spring Systems

As shown in the figure, S1S_1 and S2S_2 are identical springs with spring constant K each. The oscillation frequency of the mass ' mm ' is ' ff '. If the spring S2S_2 is removed, the oscillation frequency will become

Figure

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Answer: (c)

  1. Both springs act on mm (one stretched, one compressed), so keff=2Kk_{eff} = 2K and f=12π2Kmf = \frac{1}{2\pi}\sqrt{\frac{2K}{m}}.
  2. With S2S_2 removed, f′=12πKm=f2f' = \frac{1}{2\pi}\sqrt{\frac{K}{m}} = \frac{f}{\sqrt{2}}
  3. So the answer is (c).
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