Small oscillations about equilibrium

[Knissp]

Homework Statement

A rod of length L and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end, both pulling in opposite directions. The springs have spring constant k, and at equilibrium their pull is perpendicular to the rod. Find the frequency of small oscillations about the equilibrium position.

Homework Equations

\tau = I \frac{d^2 \theta}{dt^2}

The Attempt at a Solution

I define counterclockwise angular displacement to have positive angle.
If the position of the rod is given a slight initial perturbation \theta, then the torques due to spring 1 (on top), spring 2 (at midpoint), and gravity are given by:

\tau_g = mg\frac{L}{2}sin(\theta)

\tau_1 = -k(L sin(\theta)) L

\tau_2 = -k(\frac{L}{2} sin(\theta)) \frac{L}{2}

The moment of inertia I of a rod about one end is I = \frac{mL^2}{3}.

Then \tau = mg\frac{L}{2}sin(\theta) - k(L sin(\theta)) L - k(\frac{L}{2} sin(\theta))\frac{L}{2} = \frac{mL^2}{3} \frac{d^2 \theta}{dt^2}

Using small the angle approximation sin(\theta)=\theta,

mg\frac{L}{2}\theta - kL^2\theta - k\frac{L^2}{4} \theta = \frac{mL^2}{3} \frac{d^2 \theta}{dt^2}

(\frac{3g}{2L} - \frac{3k}{mL} - \frac{3k}{4m}) \theta = \frac{d^2 \theta}{dt^2}

Is any of that right?

Alternatively, I know that for forces, given a potential energy function U(x), the frequency
of small oscillations is given by \omega = \sqrt{\frac{U''(x)}{m}}. Perhaps a similar technique can be applied to angles?

U(\theta) = \frac{1}{2}k x_1^2 + \frac{1}{2}k x_2^2 + mg\Delta h

x_1 = L sin(\theta)

x_2 = \frac{L}{2} sin(\theta)

\Delta h = h (1-cos(\theta))

Then using small angle approximations,

U(\theta) = \frac{1}{2}k (L \theta)^2 + \frac{1}{2}k (\frac{L}{2}\theta)^2

U(\theta) = \frac{k}{2} L^2 \theta^2 + \frac{k}{2} \frac{L^2}{4}\theta^2

U(\theta) = \frac{k}{2} (L^2 \theta^2 + \frac{L^2}{4}\theta^2)

U(\theta) = \frac{k}{2} (\frac{5}{4}L^2 \theta^2)

U(\theta) = \frac{5k}{8}L^2 \theta^2

U''(\theta) = \frac{5k}{4}L^2

Then \omega = \sqrt{\frac{\frac{5k}{4}L^2}{m}}

=\frac{L}{2}\sqrt{\frac{5k}{m}}

So, I'm not sure which technique is valid or how to continue from here. Any help would be greatly appreciated. Thank you.

 

[Redbelly98]

The Attempt at a Solution

I define counterclockwise angular displacement to have positive angle.
If the position of the rod is given a slight initial perturbation \theta, then the torques due to spring 1 (on top), spring 2 (at midpoint), and gravity are given by:

\tau_g = mg\frac{L}{2}sin(\theta)

\tau_1 = -k(L sin(\theta)) L

\tau_2 = -k(\frac{L}{2} sin(\theta)) \frac{L}{2}

The moment of inertia I of a rod about one end is I = \frac{mL^2}{3}.

Then \tau = mg\frac{L}{2}sin(\theta) - k(L sin(\theta)) L - k(\frac{L}{2} sin(\theta))\frac{L}{2} = \frac{mL^2}{3} \frac{d^2 \theta}{dt^2}

Using small the angle approximation sin(\theta)=\theta,

mg\frac{L}{2}\theta - kL^2\theta - k\frac{L^2}{4} \theta = \frac{mL^2}{3} \frac{d^2 \theta}{dt^2}

(\frac{3g}{2L} - \frac{3k}{mL} - \frac{3k}{4m}) \theta = \frac{d^2 \theta}{dt^2}

Is any of that right?

Yes, though the middle term on the LHS should not have an L in the denominator. Perhaps just a typo?

From here, you can get ω simply by inspection of the differential equation.

 

[Math Jeans]

Your period is contained in the term before your \theta.

You should make the quick correction suggested, and see if you can get equivalent answers. If you can, then you're fine for either. If you cannot, try using Lagrange's equations to check a third method.

 

[Redbelly98]

Alternatively, I know that for forces, given a potential energy function U(x), the frequency
of small oscillations is given by \omega = \sqrt{\frac{U''(x)}{m}}. Perhaps a similar technique can be applied to angles?

U(\theta) = \frac{1}{2}k x_1^2 + \frac{1}{2}k x_2^2 + mg\Delta h

x_1 = L sin(\theta)

x_2 = \frac{L}{2} sin(\theta)

\Delta h = h (1-cos(\theta))

Then using small angle approximations,

U(\theta) = \frac{1}{2}k (L \theta)^2 + \frac{1}{2}k (\frac{L}{2}\theta)^2.

Just reread your post more carefully, now I realize that you are trying a different method here.

Note that terms proportional to θ² will contribute to U''(θ). For that reason we should approximate cos(θ) as (1-θ²/2), rather than simply 1, and include the gravitational term in U(θ).

 

[Knissp]

Er right that was a typo. Thanks for the help!

 

[eptheta]

I know this is 2 years old, but I did the same sum just now and got almost exactly what the OP has, except for one thing:
\tau_g=\vec{r} \times \vec {mg}
=\frac{L}{2}\hat{r} \times mg (-\hat{j})
=\frac{mgL}{2}([cos\theta\hat{i} +sin\theta\hat{j}] \times -\hat{j})
=\frac{mgL}{2}cos\theta

and not \frac{mgL}{2}sin\theta
so after small angle approximation (sinθ≈θ, cosθ≈1)

mg\frac{L}{2} - kL^2\theta - k\frac{L^2}{4} \theta = \frac{mL^2}{3} \frac{d^2 \theta}{dt^2}

As far as I can see, this isn't the simple harmonic form \frac{d^2 \theta}{dt^2}=k\theta
It is still solvable using what I know of differential equations, but would that be the correct frequency of oscillation or am I missing something ?

Thank you.

 

[Redbelly98]

I know this is 2 years old, but I did the same sum just now and got almost exactly what the OP has, except for one thing:
\tau_g=\vec{r} \times \vec {mg}
=\frac{L}{2}\hat{r} \times mg (-\hat{j})
=\frac{mgL}{2}([cos\theta\hat{i} +sin\theta\hat{j}] \times -\hat{j})
=\frac{mgL}{2}cos\theta

and not \frac{mgL}{2}sin\theta

The OP was solving for a vertical rod, while you appear to be solving for a horizontal rod.

so after small angle approximation (sinθ≈θ, cosθ≈1)

mg\frac{L}{2} - kL^2\theta - k\frac{L^2}{4} \theta = \frac{mL^2}{3} \frac{d^2 \theta}{dt^2}

As far as I can see, this isn't the simple harmonic form \frac{d^2 \theta}{dt^2}=k\theta
It is still solvable using what I know of differential equations, but would that be the correct frequency of oscillation or am I missing something ?

Thank you.

I suggest (1) combine the like terms (with θ in them) and simplify, and then (2) solve the equation for \frac{d^2 \theta}{dt^2}.

Let's see what you get after doing that, then I'll take another look at it.

 

[eptheta]

Actually, the rod I'm talking about is also vertical. Here's an image:

The rod is pivoted at O and torque calculations were done about this point.
As far as I can see, the weight still contributes a cosθ component to torque.

The equation simplifies to \ddot{\theta} + \frac{15k}{4m}\theta - \frac{3g}{2L} = 0

Whose solution is:
\theta(t)= \frac{2mg}{5kL} + c_{1}sin(\sqrt{\frac{15k}{m}}t)+ c_{2}cos(\sqrt{\frac{15k}{m}}t)

Is this form still simple harmonic ?
I can clearly see that this function does in fact have a period:
ω = \sqrt{\frac{15k}{m}}
and that the only difference between this and the general SHM solution is the addition of the constant \frac{2mg}{5kL} so the resulting periodic wave is shifted up the y-axis a bit...

 

[eptheta]

Anyone ?