How Does a Perturbed Electric Dipole Exhibit Simple Harmonic Motion?

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CAF123
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Homework Statement


A molecule with electric dipole moment ##\underline{p}## is initially aligned in an electric field ##\underline{E}## . If this molecule is perturbated from its equilbrium position by a small angle, show that it will perform simple harmonic motion.

Calculate the frequency of this motion in terms of p, E and I

The Attempt at a Solution


What I did first was write the angular EOM for the dipole: Consider it perturbed at some angle ##\theta##. This gives a torque due to the force by the electric field about the centre of the dipole. I could also consider the torque due to gravity, however, I took gravity to be acting through the centre of mass of the dipole and since I take the torques about the centre, I can ignore it. My final expression is $$I \alpha = -pE\sin \theta\,\Rightarrow\,I \alpha = -pE \theta,$$ if I take the dipole moment p = dq, sinθ ≈ θ for small θ and the -ve because the torque acts to lower θ.

Is it enough to say from here that since this is in the form ##\tau = -k \theta,## with ##k = pE##, then the motion is simple harmonic?

If so, I can say that $$T = 2\pi \sqrt{\frac{I}{k}}\,\Rightarrow\,f = \frac{1}{2 \pi}\sqrt{\frac{k}{I}}\,\Rightarrow\,f = \frac{1}{2\pi} \sqrt{\frac{pE}{I}}.$$ The dimensions check but have I made appropriate assumptions etc and is it okay to state we have the form ##\tau = -k\theta## so SHM applies?
Many thanks
 
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CAF123 said:

Homework Statement


A molecule with electric dipole moment ##\underline{p}## is initially aligned in an electric field ##\underline{E}## . If this molecule is perturbated from its equilbrium position by a small angle, show that it will perform simple harmonic motion.

Calculate the frequency of this motion in terms of p, E and I

The Attempt at a Solution


What I did first was write the angular EOM for the dipole: Consider it perturbed at some angle ##\theta##. This gives a torque due to the force by the electric field about the centre of the dipole. I could also consider the torque due to gravity, however, I took gravity to be acting through the centre of mass of the dipole and since I take the torques about the centre, I can ignore it. My final expression is $$I \alpha = -pE\sin \theta\,\Rightarrow\,I \alpha = -pE \theta,$$ if I take the dipole moment p = dq, sinθ ≈ θ for small θ and the -ve because the torque acts to lower θ.

Is it enough to say from here that since this is in the form ##\tau = -k \theta,## with ##k = pE##, then the motion is simple harmonic?

If so, I can say that $$T = 2\pi \sqrt{\frac{I}{k}}\,\Rightarrow\,f = \frac{1}{2 \pi}\sqrt{\frac{k}{I}}\,\Rightarrow\,f = \frac{1}{2\pi} \sqrt{\frac{pE}{I}}.$$ The dimensions check but have I made appropriate assumptions etc and is it okay to state we have the form ##\tau = -k\theta## so SHM applies?
Many thanks
The equation, [itex]\displaystyle \ \ I \alpha = -pE \theta\,, \[/itex] is enough to show Simple Harmonic Motion.

It's equivalent to the differential equation, [itex]\displaystyle \ \ I \frac{d^2\theta}{dt^2} = -pE \theta\ .[/itex]
 
SammyS said:
The equation, [itex]\displaystyle \ \ I \alpha = -pE \theta\,, \[/itex] is enough to show Simple Harmonic Motion.

It's equivalent to the differential equation, [itex]\displaystyle \ \ I \frac{d^2\theta}{dt^2} = -pE \theta\ .[/itex]

Thanks SammyS.
I realized I could express my eqn in the form $$\ddot{\theta} + \frac{pE}{I} \theta = 0,$$ which obviously has sin and cos as solutions.

One further question I have is that when I solve this eqn I get $$\theta = A\cos \left(\sqrt{\frac{pE}{I}}\right)t + B\sin \left(\sqrt{\frac{pE}{I}}\right)t $$,

How do I know that ##\omega## (angular freq)is necessarily the argument of sin and cos? I seem to be taking it for granted.
 
CAF123 said:
Thanks SammyS.
I realized I could express my eqn in the form $$\ddot{\theta} + \frac{pE}{I} \theta = 0,$$ which obviously has sin and cos as solutions.

One further question I have is that when I solve this eqn I get $$\theta = A\cos \left(\sqrt{\frac{pE}{I}}\right)t + B\sin \left(\sqrt{\frac{pE}{I}}\right)t\ ,$$
How do I know that ##\omega## (angular freq)is necessarily the argument of sin and cos? I seem to be taking it for granted.
What is the period of [itex]\displaystyle \ \cos \left(\sqrt{\frac{pE}{I}}\ t\right)\ ?[/itex]
 
SammyS said:
What is the period of [itex]\displaystyle \ \cos \left(\sqrt{\frac{pE}{I}}\ t\right)\ ?[/itex]
It has period $$\frac{2\pi}{\left(\sqrt{\frac{pE}{I}}\right)}$$ I see how the result follows. Thanks again.