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Homework Help: Simple harmonic motion - please help

  1. Nov 26, 2008 #1
    [tex]\pi[/tex]Hi everyone,

    I hope you can help me out with the following question. I am mainly confused about the sign

    1. The problem statement, all variables and given/known data
    A block in SHM with T= 4.0 s and A=0.1 m. How long does it take the object to move from x=0 m to x= 0.06 m.

    2. Relevant equations
    x=Acos(wt + [tex]\phi[/tex])

    where T is period, w is angular frequency, phi is phase angle and A is amplitude.

    3. The attempt at a solution

    Using eq2, i solved for w, wich is [tex]\pi[/tex]/2.
    Then I set eq1 equal to 0. The value of phi is thus + or - [tex]\pi[/tex]/2. Since the object is assumed to be moving to the right (as it would have to if it wants to go from 0 to 0.06 m),. then we take the negative phase angle. Then, back to eq1, I put

    0.06=0.1cos(([tex]\pi[/tex]/2)t - [tex]\pi[/tex]/2)

    NOW, this is were my problem is.
    Apparently, there are then two possible answers,

    either 0.927 = ([tex]\pi[/tex]/2)t - [tex]\pi[/tex]/2, which gives 1.59 s
    or - 0.927 = ([tex]\pi[/tex]/2)t - [tex]\pi[/tex]/2. which gives 0.41 s

    Firstly, I dont understand when are we supposed to have a + and - option. Secondly, I dont trust the answer given, which is 0.41 s. Please help. :confused:
  2. jcsd
  3. Nov 27, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Note that you can also use:
    x=Asin(wt + [tex]\phi[/tex])
    Which would eliminate the annoying phase factor, since x = 0 at t = 0:

    Realize that you want the first solution after t = 0. Since t = 0 corresponds to θ = -π/2 = -1.57 radians, θ = -0.927 radians is the first solution. The +0.927 solution corresponds to the block being at x = 0.06 but going to the left. (Since the block goes back and forth, there are an infinite number of times that it's at x = 0.06; you want the first time it reaches that point.)
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