Simple pendulum in an electric field

[Apphysicist]

Homework Statement

A simple pendulum of length (1.0m) and mass (5.0*10^-3 kg) is placed in a uniform electric field that is directed vertically upward. The bob has a charge (-8*10^-6 C). The period of the pendulum is (1.2 s). What are the magnitude and direction of the electric field?

Homework Equations

T=2*Pi*sqrt(L/g)

F=q*E

Maybe F=ma

The Attempt at a Solution

I'm assuming the question was just being redundant since it provided the direction. Lacking a great deal of knowledge about pendulums, I at first assumed that g in the period equation could be replaced by a generic acceleration, then using the fact that the net vertical force (ignoring tension since if g were the net acceleration under regular circumstances, it would also ignore tension) is equal to the sum of gravity and coulomb force, go through a bunch of algebra to solve for the magnitude of E. I don't know if that's correct...

Otherwise, I'd imagine writing out some kind of DE would make it very complex. I could just use a few hints as to what I should do, or maybe some kind of primer on pendulums in general. Thank you very much.

 

[kuruman]

I am not sure why you are asked to find the direction of the electric when you are told that it is directed "vertically upward". Setting that aside, yes you can calculate the effective acceleration of gravity, i.e. the vector sum of 9.81 m/s² down and the acceleration due to the electric field (qE/m in what direction?) and from that deduce the magnitude of the electric field.

 

[Apphysicist]

I am not sure why you are asked to find the direction of the electric when you are told that it is directed "vertically upward". Setting that aside, yes you can calculate the effective acceleration of gravity, i.e. the vector sum of 9.81 m/s² down and the acceleration due to the electric field (qE/m in what direction?) and from that deduce the magnitude of the electric field.

So it's really as simple as the algebraic method I did at first? g in the period equation is just "effective gravity?"

I had come up with something on the order of 10⁴ N/C, which seemed like a fairly reasonable strength electric field. I appreciate your response, since I was about ready to resign myself to not getting a response. heh

 

[kuruman]

So it's really as simple as the algebraic method I did at first? g in the period equation is just "effective gravity?"

Yes, it's that simple. The pendulum bob cannot tell that the force it experiences is just gravity or the vector sum of two or more forces.

 

[rwisz]

I would approach this problem by first considering the forces acting on the pendulum bob. We have the force of gravity pulling the bob down, and the electric force pulling it up. Using Newton's second law, we can write the equation of motion for the pendulum:

ma = -mg + qE

Where m is the mass of the bob, a is its acceleration, g is the acceleration due to gravity, q is the charge on the bob, and E is the electric field.

Next, we can use the small angle approximation to simplify the equation of motion. This approximation assumes that the angle of displacement of the pendulum bob from its equilibrium position is small, and allows us to use the equation T=2π√(L/g) to calculate the period, where L is the length of the pendulum.

Using this equation, we can substitute in the value for the period given in the problem (1.2 s) and solve for the magnitude of the electric field:

T=2π√(L/g) = 1.2 s

Solving for g, we get g = (2π/L)^2. Substituting this into our equation of motion and rearranging, we get:

a = -g + (q/m)E

Now, we can use the fact that the acceleration is equal to the second derivative of the displacement with respect to time (a=d^2x/dt^2) to get:

d^2x/dt^2 = -g + (q/m)E

This is a second order differential equation, and can be solved using standard techniques. Solving for the displacement, we get:

x(t) = A*sin(ωt + φ)

Where A is the amplitude of the motion, ω is the angular frequency, and φ is the phase angle. Substituting this into our equation of motion, we get:

Aω^2*sin(ωt + φ) = -g + (q/m)E

Comparing coefficients of sin(ωt + φ) on both sides, we get:

Aω^2 = (q/m)E

Substituting in our values for A, ω, and m, we can solve for the magnitude of the electric field:

E = (mω^2)/q * A = (0.005 kg * (2π/1.2 s)^2)/(-8