Question

Show that an electron in a one-dimensional electrical potential, \(V(x)=A x^{2},\) where the constant \(A\) is a positive real number, will execute simple harmonic motion about the origin. What is the period of that motion?

Short answer

Answer: The period of the electron's simple harmonic motion is given by the expression T = 2π √(m/2A), where m is the mass of the electron and A is a positive constant.

Step-by-step solution

Derive the force experienced by the electron

The force experienced by the electron is given by the negative gradient of the potential, which is: F(x) = -dV(x)/dx = -2Ax

Determine the equation of motion for the electron

Now, using Newton's second law of motion, F = ma, we can write the equation of motion for the electron as: ma = -2Ax where m is the mass of the electron and a is its acceleration.

Show that the motion is simple harmonic

The motion will be simple harmonic if the acceleration (a) is proportional to the displacement (x) and has the opposite direction. We can rewrite the equation of motion as: a = (-2A/m)x Since A and m are constants, this equation satisfies the condition for simple harmonic motion.

Find the angular frequency of the motion

We can relate the angular frequency (ω) to the constants in the equation of motion as follows: ω^2 = 2A/m

Find the period of the motion

The period T of the motion is given by: T = 2π/ω Using the expression for ω^2 from step 4, we find: T = 2π √(m/2A) The electron undergoes simple harmonic motion with a period of T = 2π √(m/2A).