MHT CET · Physics · Oscillations
As shown in the figure, \(S_1\) and \(S_2\) are identical springs with spring constant \(K\) each. The oscillation frequency of the mass ' \(m\) ' is ' \(f\) '. If the spring \(S_2\) is removed, the oscillation frequency will become
- A f
- B 2 f
- C \(\frac{\mathrm{f}}{\sqrt{2}}\)
- D \(\sqrt{2} \cdot \mathrm{f}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{f}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Springs in parallel: \(K_{eq1} = K + K = 2K\) Initial frequency: \(f = \frac{1}{2\pi} \sqrt{\frac{K_{eq1}}{m}} = \frac{1}{2\pi} \sqrt{\frac{2K}{m}}\)
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