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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.

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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

Hootenanny

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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?

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Thank you, it seems like a nice place :)Welcome to PF,

I have these equationsWhy don't you post what you have thus far?

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.

It's 0, but I'm still left with two unknown variables, [tex]\phi[/tex] and A, am I not?HINT: What is the velocity of the particle when it is at it's amplitude?

edit: why does omega keep showing up as superscript?

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Hootenanny

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

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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.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?

- #6

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*?

Which values of [itex]\left(\omega t + \phi\right)[/itex] correspond to the particle being at

Last edited:

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

- #8

alphysicist

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

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but I don't know neither k nor m

- #10

alphysicist

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

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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).

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

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(Sorry I couldn't write this in pretty-print. I'm new here.)

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