Calculating A in SHM from x, v & a

[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.

 

[Hootenanny]

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?

 

[mudkip9001]

Welcome to PF,

Thank you, it seems like a nice place :)

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

I have these equations

x=Acos(\omegat+\phi)
v=-A\omegasin(\omegat + \phi)
a=-\omega^{2}cos(\omegat+\phi) =\omegax

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

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

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?

 

[Hootenanny]

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

 

[mudkip9001]

Sorry, I should have typed up the question from the start:

You are watching an object in SHM. When the object is displaced 0.600m to the right of its equilibrium position, it has a velocity of 2.20 m/s to the right and an acceleration of 8.40 m/s^{2} to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

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.

 

[Hootenanny]

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?

 

[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?

 

[alphysicist]

Hi mudkip9001,

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

 

[mudkip9001]

but I don't know neither k nor m

 

[alphysicist]

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

 

[alphysicist]

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

 

[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).

Thanks, I never thought of that. However, I'm still not getting the same answer as the book:

\frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} = \frac{1}{2}kA^{2}

solving for a gives me A=\sqrt{x^{2}+\frac{m}{k}v^{2}}

\frac{-x}{a}=\frac{m}{k}=0.0714

so
A=\sqrt{0.6^{2}+0.0714*2.2^{2}}=0.840m

but the book says 0.240m, where did I go wrong?

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

 

[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.)