A body performs S.H.M. under the action of force ' F₁ ' with period ' T₁ ' second. If the force is changed to ' F₂ ' it performs S.H.M. with period ' T₂ ' second. If both forces ' F₁ ' and ' F₂ ' act simultaneously in the same direction on the body, the period in second will be (a) (T₁+T₂)/(T₁T₂) (b) (T₁²+T₂²)/(T₁T₂) (c) (T₁T₂)/(√(T₁²+T₂²)) (d) (T₁T₂)/(T₁+T₂)

Answer: (c). Forces add: k = k₁ + k₂, and k ∝ 1/T². 1/T² = (1)/(T₁²) + (1)/(T₂²) T = (T₁T₂)/(√(T₁²+T₂²)), so the answer is (c).

A body performs S.H.M. under the action of force ' F₁ ' with period '… | MHT CET 2021 (20 Sep, Shift 2)

Q73MHT CET 2021 · 20 SEP · SHIFT 2Medium

Spring Systems

A body performs S.H.M. under the action of force ' $F_1$ ' with period ' $T_1$ ' second. If the force is changed to ' $F_2$ ' it performs S.H.M. with period ' $T_2$ ' second. If both forces ' $F_1$ ' and ' $F_2$ ' act simultaneously in the same direction on the body, the period in second will be

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Answer: (c)

Forces add: $k = k_1 + k_2$ , and $k \propto \frac{1}{T^2}$ .
$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$
$T = \frac{T_1T_2}{\sqrt{T_1^2+T_2^2}}$ , so the answer is (c).

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Spring Systems question from MHT CET 2021 (20 Sep, Shift 2)

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