How Does a Perturbed Electric Dipole Exhibit Simple Harmonic Motion?

AI Thread Summary
A molecule with an electric dipole moment exhibits simple harmonic motion when perturbed from its equilibrium position in an electric field. The torque acting on the dipole leads to the equation of motion, which can be expressed as Iα = -pEθ, indicating that the motion is harmonic. The frequency of this motion can be calculated as f = (1/2π)√(pE/I), confirming the relationship between the dipole moment, electric field, and moment of inertia. The discussion also clarifies that the solutions to the differential equation governing this motion involve sine and cosine functions, with the angular frequency being the argument of these functions. The period of the motion is derived as 2π/√(pE/I), providing a complete understanding of the system's dynamics.
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, \displaystyle \ \ I \alpha = -pE \theta\,, \ is enough to show Simple Harmonic Motion.

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

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

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 \displaystyle \ \cos \left(\sqrt{\frac{pE}{I}}\ t\right)\ ?
 
SammyS said:
What is the period of \displaystyle \ \cos \left(\sqrt{\frac{pE}{I}}\ t\right)\ ?
It has period $$\frac{2\pi}{\left(\sqrt{\frac{pE}{I}}\right)}$$ I see how the result follows. Thanks again.
 
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