How can I calculate the period of a dipole's S.H.O in a uniform electric field?

[radiogaga35]

(This question doesn't apply to a specific problem, hence I'm not using the template.)

Consider a dipole, approximated as a "dumbbell": two oppositely-charged spheres (charges of equal mag.) connected by a rod (that is, the chemical bond). Suppose that I know the dipole moment. If this dipole is placed in a uniform electric field, then it would experience a net torque (but no net force).

I would like to know how I could calculate the period of the dipole's oscillations in this field? Given the electric field (vector), I know to calculate the torque (τ = p X E). From here, however, I don't know how I might use this information to calculate the period/freq. of the dipole's S.H. Oscillation. Any advice?

Thanks in advance

 

[radiogaga35]

Oh, and if it would be necessary, the mass of each "sphere" is known (or at least, it could be deduced from the elements making the molecule).

 

[chaoseverlasting]

We know \vec{T}=\vec{p}\times \vec{E}. Also, \vec{T}=I\vec{\alpha}.

This gives, I\vec{\alpha}=\vec{p}\times \vec{E}.

\alpha=\frac{d^2\theta}{dt^2}

and, \vec{p}\times \vec{E}=|\vec{p}||\vec{E}|sin\theta

Substituting, \frac{d^2\theta}{dt^2}=\frac{|\vec{p}||\vec{E}|sin\theta}{I}.

For small values of \theta, sin\theta =\theta (appox.).

\frac{d^2\theta}{dt^2}=\frac{|\vec{p}||\vec{E}|\theta}{I}This DE represents SHM, and is of the form \frac{d^2x}{dt^2}+\omega ^2x=0.

This gives \omega ^2=\frac{|\vec{p}||\vec{E}|}{I}.

Now,
\omega=2\pi \nu

This gives,
\nu=\frac{1}{2\pi}\sqrt{\frac{|\vec{p}||\vec{E}|}{I}}

which is your frequency of oscillation.

For the moment of inertia, you will need the masses of the atoms and the distance between them which can be found out from the dipole moment.

 

[radiogaga35]

Thank you very much for the comprehensive reply! Just what I was looking for. The approximation that sine(θ) = θ for a small value of θ, helped a lot (I should've thought of that...)

Am indebted!

 

[chaoseverlasting]

No prob. I am sure anyone here would have done the same.