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Calculating A in SHM from x, v & a

  • Thread starter mudkip9001
  • Start date
20
0
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.
 

Answers and Replies

Hootenanny
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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?
 
20
0
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([tex]\omega[/tex]t+[tex]\phi[/tex])
v=-A[tex]\omega[/tex]sin([tex]\omega[/tex]t + [tex]\phi[/tex])
a=-[tex]\omega[/tex][tex]^{2}[/tex]cos([tex]\omega[/tex]t+[tex]\phi[/tex]) =[tex]\omega[/tex]x

I could work out [tex]\omega[/tex] 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, [tex]\phi[/tex] and A, am I not?

edit: why does omega keep showing up as superscript?
 
Hootenanny
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Have you any initial conditions, such as whether it starts from x=A or x=0?
 
20
0
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[tex]^{2}[/tex] 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
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It seems to me that you have a system of simultaneous equations.

Which values of [itex]\left(\omega t + \phi\right)[/itex] correspond to the particle being at x=A?
 
Last edited:
20
0
when x=A, [itex]\left(\omega t + \phi\right)[/itex]=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
Homework Helper
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Hi mudkip9001,

Try conservation of energy. Equate the energy at the given point to the energy at the amplitude. What do you get?
 
20
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but I don't know neither k nor m
 
alphysicist
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You'll be able to get rid of them by using what you know about the acceleration.
 
alphysicist
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What I mean is that although you don't know either k or m, you can find the value of (k/m).
 
20
0
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:

[tex]\frac{1}{2}[/tex]kx[tex]^{2}[/tex] + [tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex] = [tex]\frac{1}{2}[/tex]kA[tex]^{2}[/tex]

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

[tex]\frac{-x}{a}[/tex]=[tex]\frac{m}{k}[/tex]=0.0714

so
A=[tex]\sqrt{0.6^{2}+0.0714*2.2^{2}}[/tex]=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.
 
1
0
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.)
 

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