1. Analysis of Force and Motion

The potential energy of the particle is given by $U(x) = k|x|$. We can determine the force acting on the particle using the relation $F = -\frac{dU}{dx}$.

• For $x > 0$, $U = kx \implies F = -k$ (Constant force directed left).
• For $x < 0$, $U = -kx \implies F = +k$ (Constant force directed right).

The force has a constant magnitude $|F| = k$ and is always directed towards the origin. This acts as a restoring force. Since the force is constant (not proportional to $x$), the acceleration is constant in magnitude in each region, unlike Simple Harmonic Motion.

The acceleration magnitude is: $$ a = \frac{|F|}{m} = \frac{k}{m} $$

2. Finding the Amplitude (Turning Point)

The particle is projected from the origin ($x=0$) with kinetic energy $K$. As it moves away, the retarding force does negative work, converting kinetic energy into potential energy. The particle stops momentarily at the extreme position $x = S$.

Using the Conservation of Mechanical Energy between the origin and the turning point:

$$ K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}} $$ $$ K + 0 = 0 + k|S| $$ $$ \Rightarrow S = \frac{K}{k} $$

3. Calculation of Time Period

The total time period $T$ consists of 4 symmetric segments: $0 \to S$, $S \to 0$, $0 \to -S$, and $-S \to 0$. Due to symmetry, the time taken for each segment is identical. Let $t_1$ be the time taken to travel from the extreme position $S$ back to the origin $0$.

Consider the motion from $x=S$ (where velocity is zero) to $x=0$.

• Initial velocity, $u = 0$
• Displacement, $\Delta x = S$
• Acceleration magnitude, $a = \frac{k}{m}$

Using the kinematic equation $s = ut + \frac{1}{2}at^2$:

$$ S = 0 + \frac{1}{2} a t_1^2 $$ $$ t_1 = \sqrt{\frac{2S}{a}} $$

Substituting the values of $S = \frac{K}{k}$ and $a = \frac{k}{m}$:

$$ t_1 = \sqrt{\frac{2(K/k)}{(k/m)}} $$ $$ t_1 = \sqrt{\frac{2Km}{k^2}} = \frac{1}{k}\sqrt{2Km} $$

4. Final Answer

The total period of the bound motion is $T = 4t_1$.

$$ T = 4 \times \frac{1}{k}\sqrt{2Km} = \frac{4}{k}\sqrt{2mK} $$