A particle of mass m is constrained to move on the x-axis. A force F acts on the particle. F always points toward the position labelled E. For example, when the particle is to the left of E , F points to the right. The magnitude of F is constant except at point E where it is zero. The system is horizontal. F is the net force acting on the particle. The particle is displaced a distance A towards left from the equilibrium position E and released from rest at t = 0. What is the period of the motion ?

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$\sqrt{\frac{2 A m}{F}}$

$4 \sqrt{\frac{2 A m}{F}}$

$2 \sqrt{\frac{2 A m}{F}}$

None

answer is A.

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Detailed Solution

T = 4t

$A = 0 + \frac{1}{2} a t^{2} A = \frac{1}{2} \frac{F}{m} t^{2} T = 4 \sqrt{\frac{2 m A}{F}}$

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