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Q.

When a particle is performing SHM of time period 3 sec, the force acting on it is F₁ for a certain displacement. When the same particle is executing SHM of time period 4 sec, the force acting is F₂ for same displacement. What will be the time period of the particle when a combined force of F₁ and F₂ produces same displacement in SHM?

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a

2.4s

b

5s

c

3.2s

d

3.5s

answer is C.

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Detailed Solution

$\begin{array}{l} F_{1} = - K_{1} x; F_{2} = - K_{2} x; F_{1} + F_{2} = - Kx; \Rightarrow K = K_{1} + K_{2} \\ \Rightarrow {mω}^{2} = {mω}_{1}^{2} + {mω}_{2}^{2} \Rightarrow ω^{2} = ω_{1}^{2} + ω_{2}^{2} \Rightarrow T = \frac{T_{1} T_{2}}{\sqrt{T_{1}^{2} + T_{2}^{2}}} = \frac{3 \times 4}{\sqrt{3^{2} + 4^{2}}} = \frac{12}{5} = 2.4 s \end{array}$

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