# 2008 us physics olympiad pendulum in electric field

1. Jan 14, 2017

### vishnu 73

1. The problem statement, all variables and given/known data

i was solving the 2008 semi final us physics olympiad paper when i got stuck on question B2 in part 2
the link takes you a pdf with questions and solution however i dont understand the soution to B2 i get part ai) but not part aii) how do you prove the motion is simple harmonic and i am not sure how they derive the period in that way

2. Relevant equations
T = 2π/ω
f = q E
V = Ed
f = mg
sin θ ≈ tan θ ≈ θ

3. The attempt at a solution
i tried to turn the coordinate system such that the equilibrium is in a vertical position but failed miserably i also tried taking a force approach and also failed i have no idea how to start please help

2. Jan 15, 2017

### Orodruin

Staff Emeritus
Please show us your attempt. If you just say that you attempted it we have no way of knowing where you went wrong.

Note that this is true only for small angles. The way the problem is stated, there is no guarantee that $\theta_0$ or $\theta$ is small. What is small in the problem is the difference $\theta - \theta_0$.

3. Jan 15, 2017

### vishnu 73

the problem is i know no way of how to start
i dont even how to set up the differential equation for this problem and i meant θ - θ0 is small not the actual angle itself
the reason why i say that is that because the force due to electric field and gravity are constant throughout the motion then why should there be a restoring force unlike the spring oscillator in which there is a restoring force thus there is SHM

Last edited: Jan 15, 2017
4. Jan 15, 2017

### Orodruin

Staff Emeritus
Well, you said that you wanted to try to write the problem in a rotated coordinate system. What do you get when you do that?

Also, the gravitational force on a standard pendulum is constant - yet it performs SHM for small angles. Do you understand why?

5. Jan 15, 2017

### vishnu 73

oh wait i forgot that while force remains the same the tangential components and the radial components of the force change let me try again give me some time

6. Jan 20, 2017

### vishnu 73

OK now i get the solution what it essentially means is that there is combined effective "gravity" pointing not directly down but at an angle of θo and then when you do the coordinate transformation such that the effective "gravity" is pointing directly down you get the equation from then on its just mathematical manipulation am i right?