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Harmonic Oscillator

  1. Nov 24, 2006 #1
    A simple harmonic oscillator consists of a block of mass 2.60 kg attached to a spring of spring constant 380 N/m. When t = 2.30 s, the position and velocity of the block are x = 0.148 m and v = 4.080 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s?

    For this problem, there was a hint online that for part A, said to use the identity sin squared (a) + cos squared (a) = 1, where a was equal to (omega x t + phi) (i'm not sure about the last greek symboL) and said to use this to relate the equation for velocity and position and solve for x subscript m. But I have no clue how to do this or any of the problem for that matter, but please help, I really need it
     
  2. jcsd
  3. Nov 24, 2006 #2

    OlderDan

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    Can you find the frequency of the oscillator from the information given? The general form for the position of an oxcillator is

    x = Asin(ωt + φ)

    The first derivative wrt time gives the velocity. That will involve a cosine function. You do not know A or φ, but since you know the x and v at a particular time, you can find them. x and ω have different dimensions, so to use the hint you will want to use the squares of either x*ω and v, or x and v/ω.
     
  4. Nov 25, 2006 #3

    andrevdh

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    As OlderDan suggests:

    The position of the oscillator contains a [tex]\sin(b)[/tex]
    while the velocity contains a [tex]\cos(b)[/tex] where [tex]b = \omega t +\phi[/tex]
    solve these two equations for the sin and the cos
    square each of these two equations and and them up
    the sum will be 1. This then gives you an equation for the amplitude of the motion.
     
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