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Simple harmonic motion and oscillation period

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data

    An object in SHM oscillates with a period of 4.0 s and an amplitude of 10 cm. How long does the object to move from x = 0.0 cm to x = 6.0 cm?

    2. Relevant equations

    x(t) = Acos(ωt +φ )

    3. The attempt at a solution

    ω = π/2
    Acosφ = 0 ⇒φ = ±π/2
    Since object is moving to the right choose φ = -π/2

    x(t) = Acos(ωt-π/2)

    My question is how why does Acos(ωt-π/2) = =Asinωt
  2. jcsd
  3. Dec 7, 2008 #2


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    Homework Helper
    Gold Member

    Acos(ωt-π/2) = =Asinωt

    This is a standard trigonometric identity. The graph of the cosine function is a sine graph shifted by [itex]\pi/2[/itex]. Thus, if you add or subtract [itex]\pi/2[/itex] from the argument of a sine function, you'll get the cosine function and vice versa.
  4. Dec 7, 2008 #3
    OK, what if i solve this equation without using the trig idenitity

    Apparently, there are then two possible answers,

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

    I know the answer is 0.41 seconds but why. Shouldnt I use the first equation because cos-1(0.06/0.01) is equal to postive 0.927
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