# 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.

Staff Emeritus
Gold Member
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($$\omega$$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.

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?

Staff Emeritus
Gold Member
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.

Staff Emeritus
Gold Member
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:
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?

Homework Helper
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

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

Homework Helper
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:grumpy:).

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.)