A particle of mass m is constrained to move o | vedclass

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(A) The particle moves under a constant force $F$ directed towards the equilibrium position $E$. This is a case of constant acceleration motion,not si

Vedclass · Accepted Answer

$4\sqrt{\frac{2Am}{F}}$

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(A) The particle moves under a constant force $F$ directed towards the equilibrium position $E$. This is a case of constant acceleration motion,not simple harmonic motion.When the particle is at a distance $A$ from $E$,its acceleration is $a = \frac{F}{m}$.The time taken to reach $E$ from the starting position is calculated using the kinematic equation $s = ut + \frac{1}{2}at^2$. Here,$s = A$,$u = 0$,and $a = \frac{F}{m}$.$A = 0 + \frac{1}{2} \left(\frac{F}{m}ight) t_1^2 \implies t_1^2 = \frac{2Am}{F} \implies t_1 = \sqrt{\frac{2Am}{F}}$.This is the time taken to travel from the extreme position to the equilibrium position $E$. Due to symmetry,the particle will travel the same distance $A$ on the other side of $E$ and return to $E$ in the same time $t_1$,and then return to the starting position in another $2t_1$.The total time period $T$ is $4 	imes t_1 = 4\sqrt{\frac{2Am}{F}}$.

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