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A particle at the end of a spring executes S.H.M. with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T then
- T⁻¹ = t1⁻¹ + t2⁻¹
- T² = t1² + t2²
- T = t1 + t2
- T⁻² = t1⁻² + t2⁻²
Correct answer: T² = t1² + t2²
Solution
For springs in series 1/k = 1/k1 + 1/k2. Since T^2 is proportional to 1/k, T^2 = t1^2 + t2^2.
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