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Simple harmonic motion problem.

  1. Aug 10, 2008 #1
    x=xocos(2pi t/T), where xo is the maximum amplitude of oscillation and T is the period of oscillation.

    Find expressions for the velocity and acceleration of a car undergoing simple harmonic motion by differentiating x.
    [Answer: a=–(2pi/T)^2(Xo)cos(2pi t/T).]

    If xo = 0.3 m and T = 3 s, what are the maximum values of velocity and acceleration?

    I do not know the expression for velocity. ANNDD I do not know (t), so even just solving for acceleration I have a problem because I face two unknowns.
  2. jcsd
  3. Aug 10, 2008 #2


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    If you found the expression for acceleration, then you should be able to find the expression for velocity.

    HINT: Take one less derivative.

    For the maximum values, you have to plug in the maximum values of cosine and sine (whichever one appears in the formula) So, the question becomes, what are the maximum values for sine and cosine? Remember, the argument of the sine or cosine shouldn't matter.
  4. Aug 10, 2008 #3
    Recall that for any oscillation, A*cos(something) or A*sin(something) that A is the maximum value that it can possibly have since cos/sin can only go as high a "1". This is why "A" is called the amplitude of the oscillation.

    Find velocity and acceleration by simply finding the first derivative of x with respect to t and the second derivative with respect to t. Whatever is out in front of those sinusoids are the max values.
  5. Aug 10, 2008 #4
    so I can disregard sine/cos since it would just be -1/1... therefore I only focus on the bginning of the equation


    And i would just plug in getting 1.3m/s2 but that is incorrect..
  6. Aug 10, 2008 #5
    and then would velocity just be:
    2(2pi/T) = Vmax
  7. Aug 10, 2008 #6


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    Try dropping the negative sign. They are probably looking for the absolute maximum acceleration.

    This is not correct. The answer should involve x_o.
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