Under the influence of force F₁ the body oscillates with a period T₁ and due to another force F₂ body oscillates with period T₂. If both forces acts simultaneously, then the resultant period is (consider displacement is same in all three cases) (a) T = √((T₁²+T₂²)/(T₁²T₂²)) (b) T = √((T₁²T₂²)/(T₁²+T₂²)) (c) T = √((T₁²)/(T₂²)) (d) T = √(T₁² + T₂²)

Answer: (b). Same displacement, so restoring forces add: k = k₁ + k₂, and k ∝ 1/T². 1/T² = (1)/(T₁²) + (1)/(T₂²) T = √((T₁²T₂²)/(T₁²+T₂²)), so the answer is (b).

Under the influence of force F₁ the body oscillates with a period T₁… | MHT CET 2023 (13 May, Shift 2)

Q55MHT CET 2023 · 13 MAY · SHIFT 2Medium

Spring Systems

Under the influence of force $F_1$ the body oscillates with a period $T_1$ and due to another force $F_2$ body oscillates with period $T_2$ . If both forces acts simultaneously, then the resultant period is (consider displacement is same in all three cases)

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Answer: (b)

Same displacement, so restoring forces add: $k = k_1 + k_2$ , and $k \propto \frac{1}{T^2}$ .
$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$
$T = \sqrt{\frac{T_1^2T_2^2}{T_1^2+T_2^2}}$ , so the answer is (b).

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Spring Systems question from MHT CET 2023 (13 May, Shift 2)

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