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Simple Harmonic Motion-Vertical Spring

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A block with mass m =7.1 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.23 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.5 m/s. The block oscillates on the spring without friction.

    3) After t = 0.37 s what is the speed of the block?
    AND
    5) At t = 0.37 s what is the magnitude of the net force on the block?



    2. Relevant equations

    v=-Asin(ωt+θ)
    A=sqrt((m*v^2)/k) [from energy conservation, used to find A]

    3. The attempt at a solution

    Tried modeling number 3 in the equation v=-Asin(ωt+θ), used A=0.689m and ω=6.528 rad/s, yet when I plug in 0.37 s for t, I do not get the right answer. Any help from here? I know I am close. Also, for number 5 i found can find acceleration at that time and then plug into Fnet=ma???
     
  2. jcsd
  3. Aug 25, 2012 #2

    TSny

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    Hello, PhysicsIzHard.
    There's something left out of this equation. Can you spot it? Also, You'll need to think carefully about the value of the phase constant θ.
    Yes.
     
    Last edited: Aug 25, 2012
  4. Aug 25, 2012 #3
    Yes, I know I forgot the ω after the A, and I kniw the phi is 0.23 m since it starts away from the origin. I still can't get the answer tho, any ideas/answers?
     
  5. Aug 25, 2012 #4

    TSny

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    phi is an angle. So, it can't have units of meters.
     
  6. Aug 25, 2012 #5
    So, how would I convert phi into radians?
     
  7. Aug 25, 2012 #6

    TSny

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    The point where the mass hangs at rest before it is set in motion is the equilibrium point of the simple harmonic motion. Let this point be x = 0 and take the direction of positive x to be downward. So, at t = 0 the mass is located at x = 0 and has a positive initial velocity. You should be able to use this information to determine the phase constant.
     
  8. Aug 25, 2012 #7
    My teacher didn't really teach me how to do that, how would I determine that?
     
  9. Aug 25, 2012 #8

    TSny

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    The general equation for the velocity is v = -ωAsin(ωt + θ), where I've adopted your original notation of using theta for the phase angle. At t = 0, we know that the mass is passing through the equilibrium point with positive velocity. Also, the speed is a maximum when passing through the equilibrium position.

    At t = 0 the velocity equation is v = -ωAsin(θ). Now, think about what the angle θ should be so that v will have it's maximum positive value.
     
  10. Aug 25, 2012 #9
    180 degrees (pi/2)? So it is going straight down?? Or 0? And why does the omega*t part of the sine disappear from that velocity equation?
     
  11. Aug 25, 2012 #10

    TSny

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    180o equal [itex]\pi[/itex] radians. But that's not the correct value. You might remember that the sine function oscillates between +1 and -1. So, v = -ωAsinθ will have its largest positive value when sinθ = -1. For what angle θ does sinθ = -1?

    We are considering the instant when t=0.
     
  12. Aug 25, 2012 #11
    I found this: The quantity φ is called the phase constant. It is determined by the initial conditions of the motion. If at t = 0 the object has its maximum displacement in the positive x-direction, then φ = 0, if it has its maximum displacement in the negative x-direction, then φ = π. If at t = 0 the particle is moving through its equilibrium position with maximum velocity in the negative x-direction then φ = π/2. The quantity ωt + φ is called the phase.

    And I used pi/2, and the answer was right. Can you explain why this??? I don't really understand how the phase angle is even derived to be these values.
     
  13. Aug 25, 2012 #12

    TSny

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    In my previous post I pointed out that at the initial time we have v = -ωAsinθ where θ is the phase angle (which is often denoted [itex]\phi[/itex] (phi)). If you want the velocity to have its maximum negative value at the initial time, then you need to have sinθ = 1. That means that θ = [itex]\pi[/itex]/2. If you want the velocity to have its maximum positive value at t = 0, then you need to have sinθ = -1. This implies θ = -[itex]\pi[/itex]/2.

    In your specific problem, you can either take upward as the positive direction or downward as the positive direction.

    If you take upward as positive, then the initial velocity is negative. So, θ = [itex]\pi[/itex]/2.

    If you take downward as positive, the the initial velocity is positive. So, θ = -[itex]\pi[/itex]/2.

    In either case, you will get the same answers for the questions (3) and (5).
     
  14. Aug 26, 2012 #13
    Yes, but why does it equal π at the maximum displacement in the negative x-direction?? I just don't understand that, I understand what you were saying.
     
  15. Aug 26, 2012 #14

    TSny

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    Well, to see that one, you should go to the equation for x: x = Acos(ωt+θ)
    Letting t = 0, we have x = Acosθ. Maximum displacement in the negative direction means that x = -A. Now, x = Acosθ will reduce to x = -A when cosθ = -1. So, θ = [itex]\pi[/itex].

    I'm afraid it's getting late for me, so I'm headed to bed. I will check back tomorrow. Cheers.
     
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