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Simple Harmonic Motion-should be easy

  1. Dec 6, 2009 #1
    A particle's position in centimeters is given by the expression x = 40 cos(45 t + 8), where t is given in seconds.

    angular frequency= 45 rad/sec
    period= .139 s
    amplitude= 40 cm

    Heres the part i cant get but seems so simple...

    where is the particle at t= .05 sec?

    what is its velocity?

    what is its accleration?

    what is the maximum speed attained by the particle?

    So what i figured on doing is just plugging .05 into my original equation x = 40 cos(45 t + 8) to solve for distance....where t= .05, distance= 39.36....which came out to be wrong for some reason...(all is in cm)
    then for the rest i just add a "-w" infront of the equation for velocity
    and a "-w^2" for acceleration and then solve when a=0, velocity has a max or min..

    any ideas?
     
  2. jcsd
  3. Dec 6, 2009 #2
    First issue, the argument of the cosine is in radians--make sure to set your calculator for radians. I got something like -27.

    See if that helps with the remaining questions which are modified by -w and w^2 as you note.
     
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