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Simple harmonic motion and time

  1. Jan 11, 2005 #1
    A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 1.70 cm and the frequency is 2.00 Hz.

    Determine the total distance traveled between t = 0 and t = 0.75 s.

    how would you solve this?

    well i can easily solve for T, w, A, etc..

    but how can i find the total distance traveled between those 2 points? is there an easy formula i am overlooking?

    i hate oscillations
  2. jcsd
  3. Jan 11, 2005 #2


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    It's the difference between two values of "x".

  4. Jan 11, 2005 #3
    i am still clueless?

    what do u mean the diff bw the two values of x?

    how can i find the value of x at a given time??

    all i know is the amplitude...but that isnt a value of x is it?
  5. Jan 11, 2005 #4


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    You know the amplitude,the frequency,u can compute the initial phase...
    You know everything...:wink:

  6. Jan 11, 2005 #5


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    What is the period of the motion? Knowing that will tell you whether the motion was all in one direction, in which case the problem is easy, or if you need to take into account one or more changes of direction.
  7. Jan 11, 2005 #6
    do i use this formula to solve for x as a function of time?
    x(t)= Acos(wt+&)

    where w=angular frequency and &= initial phase?

    in this case the t=the time right, not the period?

    and...how do i know what & is?

    when i have pi in cos or sin, do i use the radian or degree mode in my calc?

    if i use radian or degree diff answers come out
  8. Jan 13, 2005 #7
    I'm not an expert and am currently studying this now so if it dont make no sense disreguard it okay: so we have f=2 T=0.75 A=1.7

    so we could use the equation (x=distance)
    x = sin A (2π × f × t) so: sin 0.17 (2π × 2 × 0.75) = 1.59m

    SHM is always done using radians
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