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Simple Harmonic Oscillation with given equation

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A massive object of m = 5.2 kg oscillates with simple harmonic motion. Its position as a function of time varies according to the equation x(t) = 1.6sin(∏t/1.6 + ∏/6).
    a. What is the position, velocity and acceleration of the object at t = 0s?
    b. What is the kinetic energy of the object as a function of time?
    c. At what time after t=0s is the kinetic energy first at a maximum?


    2. Relevant equations

    x(t) = 1.6sin(∏t/1.6 + ∏/6).

    3. The attempt at a solution

    a. x(0) = 1.6sin(0 + ∏/6) = 0.8 m
    v(0) = ∏cos[(∏t / 1.6) + (∏/6)] = ∏cos(∏/6) = 2.72 m/s
    a(0) = - (∏^2 / 1.6) sin [(∏t / 1.6) + (∏/6)] = - 5.33 m/s^2 (the right answer should be -3.08 m/s^2)!

    b. K(t) = 1/2mv(t)^2 = 1/2m{[∏cos[(∏t / 1.6) + (∏/6)]}^2 //plugged v from part a
    2.6∏^2cos^2[(∏t / 1.6) + (∏/6)]

    c. K' = -100sin[(∏t / 1.6) + (∏/6)]cos[(∏t / 1.6) + (∏/6)].
    K'(t) = 0 at t = .26 s (But the right answer should be 1.33 s)!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 26, 2012 #2

    Curious3141

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    Homework Helper

    Remember that sin(π/6) = 1/2 NOT √3/2! :biggrin:



    For part c), it's much easier to use a double-angle formula for cosine to express the KE as a power-one expression of a single cosine term plus a constant, then use the periodicity of the cosine function to find where the KE is next at a maximum.

    I didn't check your differentiation thoroughly, but remember that if you use differentiation and set the derivative to 0 (in which case either that sine term is zero OR the cosine term is zero), you get both local minima and local maxima, so you need to figure out which is which. A little tedious.
     
    Last edited: Jan 26, 2012
  4. Jan 26, 2012 #3
    *for b. Duh! thanks :)

    *for c. I set 3pi/2 = pit/1.6 + pi/6 (cos function) and got t = 2.13 s, which is larger than the t i got when i set pit/1.6 + pi/6 = 0 (sin function), but I still somehow missed the target t = 1.33 s, just don't know where i went wrong...
     
  5. Jan 26, 2012 #4

    Curious3141

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    Remember that sin 0 = sin π = sin 2π = 0. :wink:

    When you use the '0' you get a negative time. So try the next one.

    By right you should consider the cosine part too. In that case, cos π/2 = cos 3π/2 = 0.

    But the sine term holds the answer here.

    You really have to check to make sure the KE is a minimum at that point. That involves either calculating the second derivative, or curve sketching. Tedious, tedious.

    This is why I suggested an alternative method which really is a lot easier.
     
  6. Jan 26, 2012 #5
    Thanks alot
     
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