# Calculating A in SHM from x, v & a

1. Apr 19, 2008

### mudkip9001

how can you work out the amplitude knowing only displacement, velocity and acceleration?
I have been able to work out angular velocity, frequency and period, but I can't work out amplitude without knowing either the spring constant, the mass or the phase angle.

2. Apr 19, 2008

### Hootenanny

Staff Emeritus
Welcome to PF,

Why don't you post what you have thus far?

HINT: What is the velocity of the particle when it is at it's amplitude?

3. Apr 19, 2008

### mudkip9001

Thank you, it seems like a nice place :)

I have these equations

x=Acos($$\omega$$t+$$\phi$$)
v=-A$$\omega$$sin($$\omega$$t + $$\phi$$)
a=-$$\omega$$$$^{2}$$cos($$\omega$$t+$$\phi$$) =$$\omega$$x

I could work out $$\omega$$ by re arranging the equation for a.

It's 0, but I'm still left with two unknown variables, $$\phi$$ and A, am I not?

edit: why does omega keep showing up as superscript?

4. Apr 19, 2008

### Hootenanny

Staff Emeritus
Have you any initial conditions, such as whether it starts from x=A or x=0?

5. Apr 19, 2008

### mudkip9001

Sorry, I should have typed up the question from the start:
I struggled with this for hours and concluded that the question must be wrong, but I went to the homepage of the textbook and downloaded the errata and didn't find anything.

6. Apr 19, 2008

### Hootenanny

Staff Emeritus
It seems to me that you have a system of simultaneous equations.

Which values of $\left(\omega t + \phi\right)$ correspond to the particle being at x=A?

Last edited: Apr 19, 2008
7. Apr 20, 2008

### mudkip9001

when x=A, $\left(\omega t + \phi\right)$=0

I still can't solve it though, I just end up with division of 0, can you confirm that it really is solvable?

8. Apr 20, 2008

### alphysicist

Hi mudkip9001,

Try conservation of energy. Equate the energy at the given point to the energy at the amplitude. What do you get?

9. Apr 20, 2008

### mudkip9001

but I don't know neither k nor m

10. Apr 20, 2008

### alphysicist

You'll be able to get rid of them by using what you know about the acceleration.

11. Apr 20, 2008

### alphysicist

What I mean is that although you don't know either k or m, you can find the value of (k/m).

12. Apr 20, 2008

### mudkip9001

I just wrote all this, and then figured it out myself just as I finished. But I guess I might as well post t anyway. Thanks for all your help, I'm very relieved now (but I also feel a bit dumb, since I was so sure there was something wrong with the question:grumpy:).

where I went wrong was that the question is how much further it would go, not the amplitude.

13. Dec 3, 2009

### LawnNinja

What you were missing is squaring the quantity m/k (i.e., A = \sqrt((x^2+v^2)/(x/a)^2)
(Sorry I couldn't write this in pretty-print. I'm new here.)