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 $E$. The acceleration is $a = \frac{F}{m}$.Starting from rest at distance $A$ from

Vedclass · Accepted Answer

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

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(A) The particle moves under a constant force $F$ directed towards $E$. The acceleration is $a = \frac{F}{m}$.Starting from rest at distance $A$ from $E$,the time taken to reach $E$ is found using $s = ut + \frac{1}{2}at^2$. Here $s = A$,$u = 0$,and $a = \frac{F}{m}$.$A = 0 + \frac{1}{2} \left(\frac{F}{m}\right) t_1^2 \implies t_1 = \sqrt{\frac{2Am}{F}}$.This is the time taken to travel from the extreme position to the equilibrium position $E$. Due to symmetry,the particle will travel from $E$ to the other extreme position (at distance $A$ to the right) in the same time $t_1$.Then it returns from that extreme position to $E$ in time $t_1$,and finally from $E$ back to the starting point in time $t_1$.The total time period $T = 4t_1 = 4\sqrt{\frac{2Am}{F}}$.

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