# Mass on a spring question - coordinate system?

1. Apr 4, 2008

### Sparky_

Greetings,

Regarding a mass on a spring – I know the classic differential equation is

$$m \frac {dx^2(t)}{dt^2} + B \frac {dx(t)}{dt} + kx = f(t)$$

F(t) = outside force applied
B = damping coefficient

“X” is in the vertical direction and +x direction is down.

In reading I have seen that there are some problems involving the “bar” the spring is attached oscillating.

For example – let the bar oscillate according to h(t) where h(t) is the value above or below the original position.

I’ve seen the new differential equation:

$$m \frac {dx^2(t)}{dt^2} = -k(x-h) - B \frac {dx(t)} {dt}$$

I don’t quite see why “h” is considered in -k(x-h).

Intuitively, I want to say … –k(h)

I’m thinking h is a value on the x-axis.

Is this a coordinate system problem also?

Can you help clear this up?

Thanks
-Sparky_

Last edited: Apr 4, 2008
2. Apr 4, 2008

### jasc15

If h is a constant then that can become the reference, and your diff equation will be the same as the previous one. Unless this is a non-homogenous deff eq, then that's a different story.

3. Apr 4, 2008

### Sparky_

what if h is oscillating - h(t) = Hsin(t)?

In my book they have ... -k(x-h) or (h-x) I forget.

I'm trying to understand this.

Thanks again

4. Apr 4, 2008

### tiny-tim

… spring again … !

Hi Sparky!

The kx represents the springiness times the length of the spring.

The length of the spring has to be measured from where the "other" end of the spring is.

If that other end keeps moving, according to the formula h(t), then the length of the spring is reduced from x to x - h(t).

So it becomes:
$$m \frac {dx^2(t)}{dt^2} + B \frac {dx(t)}{dt} + k(x\,-\,h(t)) = f(t)$$​

That's all!

(and, no, it's not a coordinate thing that can generally be trasnformed away … though I suppose there may be a special trick for sin(t).)

5. Apr 7, 2008

### Sparky_

Let me see if I understand -

when you say - "If that other end keeps moving, according to the formula h(t), then the length of the spring is reduced from x to x - h(t)."

your saying the spring is oscillating and then when the other end starts moving the spring length changes (increases or decreases)

If initially the spring was at rest - ?? - would h(t) set up oscillations on the mass on the end of the spring?

If x = 0 (at the origin) then there still would be a force due to k(h(t)) since h(t) combined with inertia due to the mass would result in the spring extended or contracted from rest?

If everything starts from rest and then h(t) starts the mount at the other end of the spring starts oscillating, would initially inertia play a small role and the mass at the end not move much? (initially if you move the other end I can see how the spring would not stretch or contract much.)

How is h(t) - the oscillation at the mount on the other end different from applying on external force. -

with external force:
$$m \frac {dx^2(t)}{dt^2} + B \frac {dx(t)}{dt} + kx = f(t)$$

would f(t) - be similar to moving the other end like h(t)?

with f(t) being applied, f(t) is not multiplied by k in the above equation.

Can you confirm or clear this up?

Thanks again
-Sparky_

Last edited: Apr 7, 2008
6. Apr 7, 2008

### tiny-tim

Hi Sparky!

You're missing the point … this is a coordinate thing.

Usually, coordinate things can be transformed away. If the bar is stationary (so h is zero), and we measure x from the bar, we get the original equation.

But if we move the whole apparatus y metres to the right, where y is a constant, so that the bar is at position y, then when the other end of the spring is at position x, the length of the spring is now (x - y), and so we have to replace x by (x - y) in the equation, which gives:

$$m \frac {d(x - y)^2(t)}{dt^2} + B \frac {d(x - y)(t)}{dt} + k(x - y) = f(t)\,,$$​

which of course is the same as:

$$m \frac {dx^2(t)}{dt^2} + B \frac {dx(t)}{dt} + k(x - y) = f(t)\,,$$​

and so we're back to the original x coordinate.

Putting it simply, x is a coordinate, relative to fixed space. It is not the length of the spring … that is x - y, which only looks like x because we choose to make y = 0, giving:

$$m \frac {dx^2(t)}{dt^2} + B \frac {dx(t)}{dt} + kx = f(t)\,.$$​

This is the coordinate thing … it transforms y away! And we could do that because we were free to choose our origin of coordinates (a position zero) at the bar.

But if y varies, then we cannot choose to make y = 0 and keep the same equation.

We cannot, for example, put x - y = z, and get a similar equation in z … it would involve extra dy/dt and d^2y/dt^2 terms.

We have to choose an origin of coordinates (a position zero) which is not at the bar, and so we have to leave the k(x - y) in (we're using h instead of y, of course, so it's k(x - h(t))).
Yes, you could move h to the other side, and write it:

$$m \frac {dx^2(t)}{dt^2} + B \frac {dx(t)}{dt} + kx = f(t) + kh(t)\,.$$​

7. Apr 7, 2008

### Sparky_

Thanks Tiny-Tim,

I'll digest this tonight.

-Sparky_

8. Apr 10, 2008

### Sparky_

A few more follow-up questions:

This is obvious I think, but is this because the spring is not rigid and when you move the bar by "y" the other end is not "pushed" by the same amount? There is springiness involved (hence mass on spring oscillations)

Is y (or h) not a function of time?

Should it not sill be in the 2 derivatives above.

I do agree you end up with the correct equation. I'm puzzled how the y term is simplified out of the derivatives.

9. Apr 10, 2008

### tiny-tim

No … this is geometry, not physics.

x is what we call the position of one end of the spring, y is what we call the position of the other end.

So (x - y) is the length.
No, because I said "where y is a constant".

Those equations, involving y, endeed by showing that we can transform y away, if y is constant.

Then I started again with:
And then I wrote h instead of y, so as to match the original question.

10. Apr 10, 2008

### Sparky_

It's becoming clearer.

I'll work through this.