A particle at the end of a spring executes simple harmonic motion with a period t_1 , while the corresponding period for another spring is t_2 . If th - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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A particle at the end of a spring executes simple harmonic motion with a period t_{1}, while the corresponding period for another spring is t_{2} . If the period of oscillation with the two springs in series is T, then:

Solution & Step-by-step Explanation

For a single spring, T = 2 π \frac{m}{k} ⟹ k = \frac{4 π ^{2} m}{T ^{2}} .When two springs are in series, the equivalent spring constant k_{e q} is: \frac{1}{k _{e q}} = \frac{1}{k _{1}} + \frac{1}{k _{2}} Substituting k = \frac{C}{t ^{2}} : \frac{T ^{2}}{4 π ^{2} m} = \frac{t _{1}^{2}}{4 π ^{2} m} + \frac{t _{2}^{2}}{4 π ^{2} m} T^{2} = t_{1}^{2} + t_{2}^{2}

Try it yourself before checking the explanation above.

A particle at the end of a spring executes simple harmonic motion with a period t_{1}, while the corresponding period for another spring is t_{2} . If the period of oscillation with the two springs in series is T, then:

T = t_{1} + t_{2}

T^{2} = t_{1}^{2} + t_{2}^{2}

T^{- 1} = t_{1}^{- 1} + t_{2}^{- 1}

T^{- 2} = t_{1}^{- 2} + t_{2}^{- 2}

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