A block having mass m and charge +Q is connected to an insulating spring having a force constant k. The block lies on a frictionless, insulating, horizontal track, and the system is immersed in a uniform electric field of magnitude E directed as shown in the figure below. The block is released from rest when the spring is unstretched (at x = 0). We wish to show that the ensuing motion of the block is simple harmonic.
(a) Consider the system of the block, the spring, and the electric field. Is this system isolated or nonisolated?
(b) What kinds of potential energy exist within this system? (Select all that apply.)
(c) Consider the instant the block is released from rest to be the initial configuration of the system. The final configuration is when the block momentarily comes to rest again. What is the value of x when the block comes to rest momentarily? (Use any variable or symbol stated above along with the following as necessary: π.)
(d) At some value of x we will call x = x0, the block has zero net force on it. What analysis model describes the particle in this situation?
(e) What is the value of x0? (Use any variable or symbol stated above along with the following as necessary: π.)
(f) Define a new coordinate system x' such that x' = x − x0. Show that x' satisfies a differential equation for simple harmonic motion. (Do this on paper. Your instructor may ask you to turn in this work.)
(g) Find the period of the simple harmonic motion. (Use any variable or symbol stated above along with the following as necessary: π.)
(h) How does the period depend on the electric field magnitude?
For part C, use the conservation of energy to set up an equation:
Ki + Ui = Kf +Uf
and substitute in corresponding components:
0 + QVi = 0 + (1/2 * k*x^2 + QVf)
so simplify to get
1/2 * k*x^2 +Q*E*x
Now solve for x:
x = (2*Q*E) / k
For part D the answer is particle in equilibrium because there is zero net force on it.
Part E, use conclusion from part D, to solve:
The two forces are the spring and the electric force, and we know that
Fs = -kx and Fe = EQ
Fs + Fe = 0
so Fs = - Fe
substitute in values and solve for x:
-kx = -EQ
and so x = EQ / k
Part G, the equation for tension is T = 2pi / angular frequency
and angular frequency = squareroot of (k/m)
so T = 2pi / squareroot of (k/m)
simplifying you get that
T = 2pi * squareroot of (m/k)