- #1

mudkip9001

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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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- Thread starter mudkip9001
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- #1

mudkip9001

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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?

- #3

mudkip9001

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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.

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?

- #4

Hootenanny

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

- #5

mudkip9001

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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.

- #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:

- #7

mudkip9001

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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?

- #9

mudkip9001

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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.

- #11

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).

- #12

mudkip9001

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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.

- #13

LawnNinja

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

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