A simple pendulum is suspended from the ceiling of an accelerating car

[badluckmath]

Homework Statement

Show that $\omega^{2} = \frac{\sqrt{a^{2}+g^{2} }}{l}$ for small angles

Relevant Equations

$\frac{d^{2}\theta}{dt^{2}} = -gsin(\theta)/l- acos(\theta)/l$

Here is an image of the problem:

The problem consist in finding the moviment equation for the pendulum using Lagrangian and Hamiltonian equations.
I managed to get the equations , which are shown insed the blue box:

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Using the hamilton equations, i finally got that the equilibrium angle $\theta_{e}$ : $$\theta_{e} = \tan^{-1}(\frac{-a}{g})$$m which is the angle where $\frac{d^{2}\theta}{dt^{2}} =0$.

Now, i got stuck. I tried to solve an EDO using the small angles aproximation, but it doesn't seems to lead me anywhere, because i don't really have information for the initial values.

 

[jbriggs444]

Have you considered switching to a coordinate system where the $\theta$ = 0 at the equilibrium point?

 

[badluckmath]

Just found out that i made this post on introductory physics, perhaps it should be advanced.
But, how this could help me find the answer?

 

[jbriggs444]

Just found out that i made this post on introductory physics, perhaps it should be advanced.

It has a differential equation in it, but we can tackle that just fine here.

But, how this could help me find the answer?

Personally, I am not comfortable with the Hamiltonian formalism. So I'll be keeping this Newtonian.

If you shift to a frame of reference that accelerates with the rail car then you will have a leftward inertial force from the acceleration and a downward force from gravity. The vector sum of these is constant and is directed down and to the left at your calculated equilibrium angle.

You also have a force from the supporting string. At equilibrium this force is equal and opposite to the net force above. Hence the equilibrium.

Now call this equilibrium angle 0. Consider a small deflection by $x$ (or $\theta$) from the equilibrium angle. What is the resulting net force (or torque)? Can you turn that into an ODE?

[I dislike algebra and long messy trig formulas. So I tend to look for geometric simplifications before starting with the algebra and trig. There is probably a useful formula for the variation in sine and cosine resulting from a small deflection from a known angle, and probably some clever algebraic trick so that almost everything cancels and magic happens. But it seems far easier to just change coordinates. I bet the algebraic trick amounts to the same thing in the end]

 

[badluckmath]

About that, the hamiltonian formalism gave me the expression for the second order differential equation on $\theta$ . But i get your point, i'll be trying to solve this for a new variable : $\theta= \theta_{e} + \mu$ and solve .
Still, i dont'r really like that way.

 

[kuruman]

Homework Statement:: Show that $\omega = \frac{\sqrt{a^{2}+g^{2} }}{l}$ for small angles

The expression you want to show is dimensionally incorrect. The right hand side has dimensions of $\mathrm{[T^{-2}]}$ when it is supposed to have dimensions of $\mathrm{[T^{-1}]}$. Maybe you meant to have $\omega^2$ on the left hand side?

 

[badluckmath]

The expression you want to show is dimensionally incorrect. The right hand side has dimensions of $\mathrm{[T^{-2}]}$ when it is supposed to have dimensions of $\mathrm{[T^{-1}]}$. Maybe you meant to have $\omega^2$ on the left hand side?

Yes. Sorry about that. I'll change.

 

[kuruman]

Still, i dont'r really like that way.

There is a simpler way you might like better that does not involve equations beyond intro physics. Suppose you are an observer inside the car. You see the pendulum hanging straight down. You know that, for a pendulum $\omega=\sqrt{g/l}$. Of course in your accelerated frame symbol $g$ in the equation does not stand for 9.8 m/s² but for something else that somehow takes into account the acceleration $a$ of the car. Let's call it the "effective acceleration of gravity". Can you find what $g_{\mathrm{eff}}$ is?

 

[Hiero]

i'll be trying to solve this for a new variable : $\theta= \theta_{e} + \mu$ and solve .
Still, i dont'r really like that way.

The basic principle is that most systems follow Hooke’s law for small displacements about (stable) equilibrium.

This is because the potential energy can be expanded in a Taylor series for small displacements about $x_0$ as $U(x_0+\Delta x) = U(x_0) + \Delta x \frac{dU}{dx}\Big \rvert_{x=x_0}+0.5 (\Delta x)^2\frac{d^2U}{dx^2}\Big \rvert_{x=x_0}+...$ where the linear term is zero if $x_0$ is a point of equilibrium and so the lowest order effect about (stable) equilibrium is a spring-like potential.

My point in saying this is that we don’t have to change coordinates like @jbriggs444 suggets (though I agree with him that it’s maybe a good idea to simplify the analysis) but we DO have to expand about equilibrium.

In other words you should have taken the Taylor series about $\theta=\theta_E$ instead of $\theta=0$

Edited to fix my careless mistake as pointed out by @etotheipi (thanks!)

 

[etotheipi]

$U(x) = U(x_0) + x\frac{dU}{dx}\Big \rvert_{x=x_0}+0.5x^2\frac{d^2U}{dx^2}\Big \rvert_{x=x_0}+...$

Small correction; when you're doing a Maclaurin series remember you're considering powers of $(x-x_0)$:$$U(x) = U(x_0) + (x-x_0) \frac{dU}{dx} (x_0) + \frac{1}{2!} (x-x_0)^2 \frac{d^2 U}{dx^2}(x_0) + \dots $$