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Homework Help: Vibrations and Wave Question

  1. May 18, 2010 #1
    I have a few questions i need some help on.

    1. The problem statement, all variables and given/known data
    The position of a particle is given by the expression [tex]x=(4.00m) cos(3.00\pit+\pi)[/tex], where x is in meters and t in seconds. Determine (a) the frequency and period of the motion.

    2. Relevant equations


    3. The attempt at a solution

    i got:






    This is incorrect according to book's answer, what is wrong???

    1. The problem statement, all variables and given/known data
    A 0.500kg object attached to a spring with a force constant of 8.00N/m vibrates in simple harmonic motion with an amplitude of 10.0cm. Calculate time it takes the object to move from x=+8.0cm (just after it is released at x=+10cm) to x=0.

    3. The attempt at a solution
    i tried to use this formula: [tex]x=Acos(kt+\theta)[/tex], however i cannot find t because there is two unknowns.

    Instead what should i use?

  2. jcsd
  3. May 18, 2010 #2


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    Homework Helper

    t does not appear in the formula!

    What is k? is it the angular frequency? If so, how is the angular frequency of the vibration related to the force constant and mass?

    If the object is released at x=10 cm which is the maximum displacement, what is the value of the phase constant theta?

  4. May 18, 2010 #3
    oops let me write it out again:

    [tex]x=Acos(\omega t + \theta)[/tex]

    this is the correct equation i used

    so i sub in values and i get two unknowns

    [tex]8 = 10cos(4t+ \theta)[/tex]
  5. May 18, 2010 #4


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    Homework Helper

    The time is measured from the instant when the object is released. The object is released when x=10. So what is the value of theta?

  6. May 19, 2010 #5
    ok so this is what i have done:

    [tex]x=Acos(\omega t + \theta)[/tex]

    x = 10cm, A = 10cm, [tex]\omega[/tex] = 4m/s t = ? [tex]\theta[/tex] = 0.93 degrees

    [tex]10 = 10cos(4t + 0.93)[/tex]

    [tex]10 = 10cos(4t) +10cos(0.93)[/tex]

    [tex]10 - 10cos(0.93) = 10cos4t[/tex]

    [tex]t = \frac{arccos\frac{10-10cos(0.93)}{10}}{4}[/tex]

    [tex]t = 0.289 s[/tex]
  7. May 19, 2010 #6
    Could anyone tell me what is the problem with my first question??
  8. May 19, 2010 #7
    [tex]T=\frac{2\pi}{3\pi}[/tex] reduces down to 2/3 which is not equal to 6.58.
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