Question

A particle of mass m can move along x-axis of co-ordinate frame in a force field of stationary sources. Potential energy 'U' of system varies with position x of the particle according to the equation U = K |x|, where k is a positive constant. If the particle is projected from the origin with a kinetic energy k', and its period of bounded motion is $\frac{N}{K}\sqrt{\frac{mk'}{2}}$, find the value of N.

Answer 1

8

Answer 2

The potential energy of the particle is given by $U(x) = K|x|$, where $K$ is a positive constant. The force acting on the particle is $F(x) = -\frac{dU}{dx}$.

For $x > 0$, $U(x) = Kx$, so $F(x) = -\frac{d(Kx)}{dx} = -K$.

For $x < 0$, $U(x) = -Kx$, so $F(x) = -\frac{d(-Kx)}{dx} = -(-K) = K$.

The force is a constant restoring force, directed towards the origin, with magnitude $K$. $F(x) = -K \text{ sgn}(x)$.

The particle is projected from the origin ($x=0$) with kinetic energy $k'$. At $x=0$, the potential energy is $U(0) = K|0| = 0$. The total energy of the particle is $E = \text{Kinetic Energy} + \text{Potential Energy}$. At $x=0$, $E = k' + U(0) = k' + 0 = k'$. Since the motion is bounded, the particle oscillates between two turning points where its velocity is zero. At the turning points, the total energy is equal to the potential energy:

$E = U(x)$ $k' = K|x|$ $|x| = \frac{k'}{K}$.

Let $A = \frac{k'}{K}$. The turning points are at $x = -A$ and $x = A$. The motion is confined to the region $-A \le x \le A$.

Let's consider the motion starting from $x=0$ with an initial velocity $v_0$. The initial kinetic energy is $k' = \frac{1}{2}mv_0^2$. So, $v_0 = \sqrt{\frac{2k'}{m}}$. We can assume the particle is initially projected in the positive x-direction, so $v(0) = v_0$.

The motion is periodic. One full period consists of the particle starting from $x=0$, moving to $x=A$, returning to $x=0$, moving to $x=-A$, and finally returning to $x=0$.

Consider the motion from $x=0$ to $x=A$. For $x > 0$, the force is $F(x) = -K$. The acceleration is $a = \frac{F}{m} = -\frac{K}{m}$. This is motion with constant acceleration. The initial velocity at $x=0$ is $v_0 = \sqrt{\frac{2k'}{m}}$. The final velocity at $x=A$ is $v=0$. We can use the kinematic equation $v = u + at$. Let $t_1$ be the time taken to go from $x=0$ to $x=A$. $0 = v_0 + (-\frac{K}{m})t_1$ $t_1 = \frac{v_0}{K/m} = \frac{m v_0}{K}$. Substitute $v_0 = \sqrt{\frac{2k'}{m}}$: $t_1 = \frac{m}{K} \sqrt{\frac{2k'}{m}} = \frac{\sqrt{m^2}}{K} \sqrt{\frac{2k'}{m}} = \frac{1}{K}\sqrt{\frac{m^2 \cdot 2k'}{m}} = \frac{\sqrt{2mk'}}{K}$.

The motion from $x=A$ back to $x=0$ is also under constant acceleration $a = -K/m$. The initial velocity at $x=A$ is 0, and the final velocity at $x=0$ will be $-v_0$ (due to symmetry, the speed will be the same as the initial speed, but in the opposite direction). Let $t_2$ be the time taken. Using $v^2 = u^2 + 2as$. The displacement is $\Delta x = 0 - A = -A$. $(-v_0)^2 = 0^2 + 2(-\frac{K}{m})(-A)$ $v_0^2 = \frac{2KA}{m}$. This is consistent with $A = k'/K$ and $v_0^2 = 2k'/m$. Using $v = u + at$: $-v_0 = 0 + (-\frac{K}{m})t_2$, so $t_2 = \frac{v_0}{K/m} = \frac{\sqrt{2mk'}}{K}$. So, the time taken from $x=A$ to $x=0$ is $t_2 = \frac{\sqrt{2mk'}}{K}$.

Now consider the motion for $x < 0$. For $x < 0$, the force is $F(x) = K$. The acceleration is $a = \frac{F}{m} = \frac{K}{m}$. This is motion with constant acceleration. The particle arrives at $x=0$ from $x=A$ with velocity $-v_0$. It then moves from $x=0$ to $x=-A$. The initial velocity at $x=0$ for this segment is $-v_0 = -\sqrt{\frac{2k'}{m}}$. The final velocity at $x=-A$ is 0. Let $t_3$ be the time taken. Using $v = u + at$: $0 = (-v_0) + (\frac{K}{m})t_3$. $t_3 = \frac{v_0}{K/m} = \frac{m v_0}{K} = \frac{\sqrt{2mk'}}{K}$. So, the time taken from $x=0$ to $x=-A$ is $t_3 = \frac{\sqrt{2mk'}}{K}$.

Finally, the motion from $x=-A$ back to $x=0$. The initial velocity at $x=-A$ is 0. The final velocity at $x=0$ will be $v_0$. The acceleration is $a = K/m$. Let $t_4$ be the time taken. Using $v = u + at$: $v_0 = 0 + (\frac{K}{m})t_4$. $t_4 = \frac{v_0}{K/m} = \frac{m v_0}{K} = \frac{\sqrt{2mk'}}{K}$. So, the time taken from $x=-A$ to $x=0$ is $t_4 = \frac{\sqrt{2mk'}}{K}$.

The total period of bounded motion is the sum of the times for these four segments: $T = t_1 + t_2 + t_3 + t_4$ $T = \frac{\sqrt{2mk'}}{K} + \frac{\sqrt{2mk'}}{K} + \frac{\sqrt{2mk'}}{K} + \frac{\sqrt{2mk'}}{K}$ $T = 4 \times \frac{\sqrt{2mk'}}{K}$.

We are given that the period of bounded motion is $T = \frac{N}{K}\sqrt{\frac{mk'}{2}}$. Let's rewrite our calculated period in the given form: $T = 4 \frac{\sqrt{2mk'}}{K} = \frac{4}{K} \sqrt{2mk'} = \frac{4}{K} \sqrt{mk' \cdot 2} = \frac{4}{K} \sqrt{mk'} \sqrt{2}$. We need the term $\sqrt{\frac{mk'}{2}}$. Let's manipulate the expression: $T = \frac{4\sqrt{2}}{K} \sqrt{mk'} = \frac{4\sqrt{2}}{K} \sqrt{mk'} \frac{\sqrt{2}}{\sqrt{2}} = \frac{4 \cdot 2}{K\sqrt{2}} \sqrt{mk'} = \frac{8}{K\sqrt{2}} \sqrt{mk'}$. This doesn't look like the required form.

Let's try another manipulation: $T = \frac{4}{K} \sqrt{2mk'} = \frac{4}{K} \sqrt{4 \cdot \frac{2mk'}{4}} = \frac{4}{K} \sqrt{4} \sqrt{\frac{2mk'}{4}} = \frac{4 \cdot 2}{K} \sqrt{\frac{mk'}{2}} = \frac{8}{K} \sqrt{\frac{mk'}{2}}$.

Comparing this with the given expression for the period $T = \frac{N}{K}\sqrt{\frac{mk'}{2}}$, we can identify $N$. $\frac{8}{K}\sqrt{\frac{mk'}{2}} = \frac{N}{K}\sqrt{\frac{mk'}{2}}$ $N = 8$.

Therefore, the value of N is 8.