Harmonic oscillator

1. May 5, 2009

Ofey

The frequency of a harmonic oscillator is (as you know)

$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

I am wondering if this equation only applies for massless harmonic oscillators (or oscillators oscillating sideways)?

The proof for the equation above is

$$\sum {F=ma}$$

$$-kx=ma$$

$$-kx=-m\omega^2x$$

$$k=m\omega^2$$

Since

$$\omega=2\pi f$$

We get the familiar equation above

$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

But as you can see we have neglected the force of gravity (if we assume that such a force exists). I solved an assignment today using this formula above. It gave the wrong answer when I just used the formula which neglected gravity. But when I went back and proved the formula "again", this time assuming the existense of gravity I ended up in the right answer. This would be a bummer, since I have always been taught that the frequency for a harmonic oscillator is:

$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

No matter what.

2. May 5, 2009

deferro

No, the m in all the equations is for the mass term; this is determinate (use a weighing balance to measure it).
The m takes care of gravity for a discrete body with sufficient mass, so you have it in the equations.
But quantum oscillators are more complicated.

3. May 5, 2009

KLoux

There is no such thing as a "massless harmonic oscillator"...

Yes, this relationship applies even in the presence of gravity. The difference in the analysis of a mass hanging on a spring in a gravitational field and a horizontal mass-spring system on a frictionless surface is the "neutral" length of the spring.

Maybe you can show us this question that you got wrong and we can try to figure out where the problem is.

-Kerry

4. May 6, 2009

Ofey

A computer used in space shuttles has to work during extreme circumstances. The computer has to work during an acceleration equivalent to 10G without breaking. The computer is tested by placing it on an oscillating horizontal plane. The amplitude of the oscillation is 0.10m. Determine the frequency of the oscillation.

The correct solution

$$x=Asin(\omega t)$$

$$v=\frac{dx}{dt}=\frac{d}{dt}(Asin(\omega t))$$

$$a=\frac{dv}{dt}=\frac{d}{dt}(A\omega cos(\omega t))=-A\omega^2sin(\omega t)$$

Since

$$sin(\omega t) \in [-1,1]$$

The maximal acceleration is given by

$$A\omega^2=A(2\pi f)^2=10g$$

Solving for frequency yields

$$f=\frac{1}{2\pi}\sqrt{\frac{10g}{A}}$$

My first attempt

This is the situation when the computer is at the lowest position in the oscillation. This is when the acceleration is at its largest. Chosing upwards to be the positive direction

$$-kx-mg=ma$$

$$-kx=m(g+a)$$

Since

$$a=10g$$

$$kx=-11mg$$

$$\frac{k}{m}=\frac{-11g}{x}$$

Placing this in the well-known equation gives

$$f=\frac{1}{2\pi}\sqrt{\frac{-11g}{x}}$$

Where $$x=-A$$

Then I tried to prove the frequency for a harmonic oscillator again, this time assuming gravitational effects.

$$\Sigma F=ma$$

(I mean it says the sum of all forces, gravity does act on the body so surely it has to be added?)

$$-kx-mg=ma$$

$$-kx=m(g-\omega^2x)$$

$$\frac{k}{m}=\frac{\omega^2x-g}{x}$$

Again, since

$$\omega=2\pi f$$

$$f=\frac{1}{2\pi}\sqrt{\frac{kx+mg}{mx}}$$

Since

$$kx=-11mg$$

$$f=\frac{1}{2\pi}\sqrt{\frac{-11mg+mg}{mx}}$$

$$f=\frac{1}{2\pi}\sqrt{\frac{-10g}{x}}$$

Since

$$x=-A$$

We get

$$f=\frac{1}{2\pi}\sqrt{\frac{10g}{A}}$$

Which is correct? What am I missing

Last edited: May 6, 2009
5. May 6, 2009

Born2bwire

If it's oscillating in a horizontal plane, why would you have a force from gravity acting along the direction of oscillation?

6. May 6, 2009

Ofey

I apologize for my english.

The computer is placed on a horizontal plane which oscillates in the vertical direction.

7. May 6, 2009

KLoux

Ofey,

It sounds like you're introducing a spring into a problem that doesn't necessarily have a spring (unless harmonic oscillator implies that it is a mass-spring system instead of something driven by a motor).

Anyway - you're adding 10g acceleration to something that is already experiencing 1g due to Earth's gravity. This will give you 11g acceleration. You should only add 9g to get the 10g that you want.

If the platform isn't moving, the object is at 1g. If the platform is accelerated upwards relative to the ground at 1g, the object is at 2g. If it accelerates downwards at 1g, the object is at 0g.

It's difficult to give a better explanation than this without a diagram showing your system, but it seems like there's probably something going on with reaction forces between the object and the platform? Anyway, as you showed with the correct solution, it is not necessary to have any additional information to solve the problem.

I hope this helps.

-Kerry

8. May 8, 2009

Ofey

One question. In wikipedia it says:

"If F* is the only force acting on the system, the system is called a simple harmonic oscillator"

But F isn't the only force if gravity also acts on the system?

*F=-kx

In all proofs of harmonic motion gravity is neglected. I don't understand why we do that. :(

9. May 8, 2009

davieddy

It is taken into account by choosing x = 0 to be the point
where the spring force and mg are equal and opposite.

Then you find the net force is -kx

10. May 8, 2009

Ofey

Now I understand. Thank you :)