A particle connected to the end of a spring executes S.H.M. with period ' T₁ '. While the corresponding period for another spring is ' T₂ '. If the period of oscillation with two springs in series is 'T', then (a) T = √(T₁² + T₂²) (b) T = √(T₂² - T₁²) (c) T = T₁ + T₂ (d) T = T₁ - T₂

Answer: (a). Series: 1/k = 1/k₁ + 1/k₂. T² = (4π² m)/(k) ∝ 1/k, so T² = T₁² + T₂². So the answer is (a) T = √(T₁²+T₂²).

A particle connected to the end of a spring executes S.H.M. with… | MHT CET 2021 (22 Sep, Shift 2)

Q72MHT CET 2021 · 22 SEP · SHIFT 2Medium

Spring Systems

A particle connected to the end of a spring executes S.H.M. with period ' $T_1$ '. While the corresponding period for another spring is ' $T_2$ '. If the period of oscillation with two springs in series is 'T', then

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Answer: (a)

Series: $\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}$ .
$T^2 = \frac{4\pi^2 m}{k} \propto \frac{1}{k}$ , so $T^2 = T_1^2 + T_2^2$ .
So the answer is (a) $T = \sqrt{T_1^2+T_2^2}$ .

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Spring Systems question from MHT CET 2021 (22 Sep, Shift 2)

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