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Finding amplitude of oscillating spring; x(t) and v(t) given

  1. Sep 24, 2016 #1

    tensor0910

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    1. The problem statement, all variables and given/known data

    At t = 0 seconds, the position of the oscillator is found to be -27.33 cm and the velocity of the oscillator is +0.744 m/s. What is the amplitude of the oscillation?





    2. Relevant equations

    position/time equation for oscillation: x(t) = Acos(wt + phi )
    for velocity: v(t) = -Vmax sin(wt + phi )



    3. The attempt at a solution

    at t = 0, the position is -.2733 m and the velocity is .744 m/s.
    This leaves us with two equations with 2 unknowns. ( I think ) this is how they should look:

    -.2733 = Acos(phi)
    .744 = -.744sin (phi )

    I unsuccessfully attempted to solve for either unknown and ended up with:

    tan(phi ) = -A/-.2733

    This is where I got stuck. Not even sure if my algebra is correct. Any/all tips are appreciated. Thank you in advance!

    Edit: Original problem coupled with question. Sorry, forum noob here :-)

    A mass of 1.4 kg is attached to a single spring and set into oscillation. The period of the spring oscillation is measured to be 4.22 seconds

     
    Last edited: Sep 24, 2016
  2. jcsd
  3. Sep 24, 2016 #2
    Hi tensor:

    I see two things that appear to be not quite right in your work.
    1. You omitted the equation v(t) = dx/dt.
    2. How did you arrive at the value of the coefficient of sin in the quoted equation?

    Hope this helps.

    Regards,
    Buzz
     
  4. Sep 24, 2016 #3

    LCKurtz

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    Please do not bold face your posts. It violates forum rules.

    If ##x = \cos(\omega t + \phi)## then ##v = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)##, so you have three unknowns ##A,~\omega,\text{ and }\phi##. Putting ##t=0## in those two gives two equations. And if you know the period is ##4.22## seconds, that should give you ##\omega##. It might be helpful to use the ##\sin^2 + \cos^2## identity in your work.
     
  5. Sep 24, 2016 #4

    tensor0910

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    thank you for the reply Buzz!

    After relooking at the problem with Buzz's tip I came up with these equations:

    -.2733 = Acos( Φ ) x(t)
    .744 = -ωAsin( Φ ) v(t)

    I solved for ω using 2π/T ( calculator in rad ) and came up with 1.488

    solved for A: A = -.2733/cosΦ

    plugged A into v(t) and came up with :

    .744 = (.40667)(tanΦ)

    isolated tan(Φ) and solved for Φ and came up with...1.070.

    But the program says its wrong!

    Does anyone have any ideas?

    Edit: solved for Amplitude using -.2733/cos (1.070.)
     
    Last edited: Sep 24, 2016
  6. Sep 24, 2016 #5

    tensor0910

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    Figured it out. Calculator needs to be back in Deg. to finish problem. Φ = 61.338. Plug into x(t)

    Thank you Buzz and LCKurtz for the help!
     
  7. Sep 25, 2016 #6

    ehild

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    The amplitude is asked, you do not need the angle.
    -0.2733 = Acos( Φ ) = x(t)
    0.744 = -ωAsin( Φ ) = v(t)
    ω=1.489
    You can write the second equation in form Asin( Φ ) = 0.744/ω. Square both this equation and the first one, add up the squares, and use @LCKurtz hint about sin2 + cos2.
     
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