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Simple harmonic motion of a machine part

  1. Nov 20, 2012 #1
    1. The problem statement, all variables and given/known data
    A machine part is undergoing SHM with a frequency of of 5hz and amplitude of 1.80cm. How long does it take the part to go from x = 0 to -1.80cm?


    2. Relevant equations
    x = Acoswt


    3. The attempt at a solution
    X is given and convert it to metres 0.018. I need to manipulate answer to get time.
    angular velocity is 2pi x 5 = 31.41. It is just an equation manipulation. could someone please show me the technique?
     
  2. jcsd
  3. Nov 20, 2012 #2
    The displacement is the negative of the amplitude.
     
  4. Nov 20, 2012 #3

    CWatters

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    If the peak amplitude is 1.8cm then asking when the displacement is -1.8 makes it rather easy. No need to even use the sin button on the calculator :-)

    Sketch displacement vs time. Mark the points 1.8m and -1.8m. If T is the period (=1/5) at what fraction of T does 1.8cm and -1.8cm occur?
     
  5. Nov 20, 2012 #4
    thanks 1/5 equals 0.2. Then divide by 4 and 0.05 is the time in seconds
     
  6. Nov 21, 2012 #5

    CWatters

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    0.05s is one valid answer but there are two possible valid right answers for this question. I recommend you give both.
     
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