Initial conditions in simple harmonic motion

  • #1
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Homework Statement:
A particle of ##100 g## moves with simple harmonic motion. When the particle is ##8 cm## away from its position of equilibrium its velocity is ##45 \frac{m}{s}## and its acceleration is ##300 \frac{m}{s^2}## opposing the motion. Find the amplitude and frequency of oscilation.
Relevant Equations:
##A=\sqrt (x_0^2 + \frac{\dot x_0 ^2}{\omega^2})##
I have the formula for amplitude ##A=\sqrt (x_0^2 + \frac{\dot x_0 ^2}{\omega^2})##.
But ##x_0## and ##\dot x_0## refers to the initial conditions, and the information that I'm given is not related to the initial conditions, or at least I'm not told so.
 

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  • #2
andrewkirk
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You can treat the numbers you were given as initial conditions. Initial conditions doesn't mean when the motion started, it means at the start of the period you are analysing.
 
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  • #3
Orodruin
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Also, your formula is a relationship between amplitude and frequency if you know the position and speed. Since you have two unknowns, you are going to need one more relation in order to solve for both.
 
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  • #4
PeroK
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Homework Statement: A particle of ##100 g## moves with simple harmonic motion. When the particle is ##8 cm## away from its position of equilibrium its velocity is ##45 \frac{m}{s}## and its acceleration is ##300 \frac{m}{s^2}## opposing the motion. Find the amplitude and frequency of oscilation.
Homework Equations: ##A=\sqrt (x_0^2 + \frac{\dot x_0 ^2}{\omega^2})##

I have the formula for amplitude ##A=\sqrt (x_0^2 + \frac{\dot x_0 ^2}{\omega^2})##.
But ##x_0## and ##\dot x_0## refers to the initial conditions, and the information that I'm given is not related to the initial conditions, or at least I'm not told so.

In what fundamental way do "initial conditions" differ from the state of the system at any other time?
 
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  • #5
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Thanks
I used the formula but I realized that I didn't have ##\omega##. So I used the equations for SHM with ##t=0##
##8=A.cos(\phi)##
##45=-A\omega sin(\phi)##
##-300=-A \omega ^2 cos (\phi)##.

So I can find ##\phi##, ##\omega## and ##A##, but I get a negative value for ##\omega## and the frequency. Is that possible?
 
  • #6
PeroK
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I used the formula but I realized that I didn't have ##\omega##. So I used the equations for SHM with ##t=0##
##8=A.cos(\phi)##
##45=-A\omega sin(\phi)##
##-300=-A \omega ^2 cos (\phi)##.

So I can find ##\phi##, ##\omega## and ##A##, but I get a negative value for ##\omega## and the frequency. Is that possible?

I'm not sure how you got that. The first and third equations ought to give youn##\omega^2##.
 
  • #7
Orodruin
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I used the formula but I realized that I didn't have ##\omega##. So I used the equations for SHM with ##t=0##
##8=A.cos(\phi)##
##45=-A\omega sin(\phi)##
##-300=-A \omega ^2 cos (\phi)##.

So I can find ##\phi##, ##\omega## and ##A##, but I get a negative value for ##\omega## and the frequency. Is that possible?
Careful with the units!
 
  • #8
kuruman
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I used the formula but I realized that I didn't have ##\omega##. So I used the equations for SHM with ##t=0##
##8=A.cos(\phi)##
##45=-A\omega sin(\phi)##
##-300=-A \omega ^2 cos (\phi)##.

So I can find ##\phi##, ##\omega## and ##A##, but I get a negative value for ##\omega## and the frequency. Is that possible?
You don't need all that to find ##\omega##. In simple harmonic motion the acceleration is always proportional to the displacement from equilibrium, the constant of proportionality being ##-\omega^2##. The minus sign is there because the acceleration vector is always in the opposite direction as the displacement vector; just strip the negative sign to get a relation between magnitudes. Here, you are given displacement and acceleration so ...
 
  • #9
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You don't need all that to find ##\omega##. In simple harmonic motion the acceleration is always proportional to the displacement from equilibrium, the constant of proportionality being ##-\omega^2##. The minus sign is there because the acceleration vector is always in the opposite direction as the displacement vector; just strip the negative sign to get a relation between magnitudes. Here, you are given displacement and acceleration so ...
Do you mean ##-k\Delta x=m.a##?
 
  • #10
kuruman
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Do you mean ##-k\Delta x=m.a##?
No. I mean what I said, ##a(t)=-\omega^2x(t)##. The second derivative of a sine or cosine is the negative of itself and the chain rule applied twice brings the square of the frequency up front. This works with the general solution ##x(t)=A\sin(\omega t)+B\cos(\omega t)##. Try it.
 
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  • #11
PeroK
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No. I mean what I said, ##a(t)=-\omega^2x(t)##. The second derivative of a sine or cosine is the negative of itself and the chain rule applied twice brings the square of the frequency up front. This works with the general solution ##x(t)=A\sin(\omega t)+B\cos(\omega t)##. Try it.

You can get that also from ##F = ma = -kx##, with ##\omega^2 = k/m##.
 
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  • #12
kuruman
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You can get that also from ##F = ma = -kx##, with ##\omega^2 = k/m##.
Indeed. I was thinking about the more general SHO equation ##\ddot x(t)=-\omega^2 x(t)~##that does not necessarily involve a spring-mass system.
 

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