Question

The potential energy U of a particle is given by U = {20 + (x - 4)^2}J. Total mechanical energy of the particle is 36 J. Select the correct alternative(s)

• A. The particle oscillates about point x = 4 m
• B. The amplitude of the particle is 4 m
• C. The kinetic energy of the particle at x = 2 m is 12 J
• D. The motion of the particle is periodic but not simple harmonic

Answer 1

Answer: (A), (B), (C)

The particle oscillates about point x = 4 m, The amplitude of the particle is 4 m, The kinetic energy of the particle at x = 2 m is 12 J

Answer 2

The potential energy of the particle is given by $U(x) = 20 + (x - 4)^2$ J. The total mechanical energy of the particle is $E = 36$ J.

The total mechanical energy is the sum of kinetic energy $K$ and potential energy $U$: $E = K + U$. So, the kinetic energy is $K(x) = E - U(x) = 36 - (20 + (x - 4)^2) = 16 - (x - 4)^2$ J.

Statement 1: The particle oscillates about point x = 4 m.

The potential energy function $U(x) = 20 + (x - 4)^2$ is a parabola opening upwards, with a minimum at $x - 4 = 0$, i.e., $x = 4$. This minimum corresponds to a stable equilibrium position. The minimum potential energy is $U_{min} = 20 + (4 - 4)^2 = 20$ J. Since the total energy $E = 36$ J is greater than the minimum potential energy $U_{min} = 20$ J, the particle is bound and will oscillate about the stable equilibrium position $x = 4$ m. Thus, statement 1 is correct.

Statement 2: The amplitude of the particle is 4 m.

The turning points of the motion are where the kinetic energy is zero ($K = 0$), which means $U(x) = E$. $20 + (x - 4)^2 = 36$. $(x - 4)^2 = 16$. $x - 4 = \pm \sqrt{16} = \pm 4$. The turning points are $x_1 = 4 - 4 = 0$ m and $x_2 = 4 + 4 = 8$ m. The equilibrium position is $x_0 = 4$ m. The amplitude of oscillation $A$ is the maximum displacement from the equilibrium position. $A = |x_2 - x_0| = |8 - 4| = 4$ m, or $A = |x_0 - x_1| = |4 - 0| = 4$ m. Thus, the amplitude of the particle is 4 m. Statement 2 is correct.

Statement 3: The kinetic energy of the particle at x = 2 m is 12 J.

The kinetic energy function is $K(x) = 16 - (x - 4)^2$. At $x = 2$ m, the kinetic energy is $K(2) = 16 - (2 - 4)^2 = 16 - (-2)^2 = 16 - 4 = 12$ J. Thus, statement 3 is correct.

Statement 4: The motion of the particle is periodic but not simple harmonic.

The potential energy function is $U(x) = (x - 4)^2 + 20$. This potential energy function is of the form $U(x) = \frac{1}{2}k(x - x_0)^2 + U_0$, which is the characteristic potential energy function for Simple Harmonic Motion (SHM), where $x_0$ is the equilibrium position and $k$ is the force constant. Comparing $U(x) = (x - 4)^2 + 20$ with the standard form, we have $\frac{1}{2}k = 1$, so $k = 2$. The equilibrium position is $x_0 = 4$ m, and $U_0 = 20$ J. The force acting on the particle is $F(x) = -\frac{dU}{dx} = -\frac{d}{dx}(20 + (x - 4)^2) = -(2(x - 4)) = -2(x - 4)$. This is a linear restoring force of the form $F(x) = -k(x - x_0)$, where $k = 2$ and $x_0 = 4$. Motion under a linear restoring force is Simple Harmonic Motion. SHM is a type of periodic motion. Therefore, the motion is simple harmonic. The statement says the motion is periodic but not simple harmonic, which is incorrect. Thus, statement 4 is incorrect.