Post a New Question


posted by .

a simple harmonic oscillator consists of a block of mass 2 kg attached to a spring of spring constant 100 N/m. when t=1 s, the position and velocity of the block are x=.129 m and v=3.415 m/s.
A) what is the amplitude
i figured that out and got sqrt(50)
B) what is the position at t=0
C) what is the velocity at t=0

  • physics -

    Get the amplitude A from
    (1/2)kx^2 + (1/2)m v^2 = (1/2) k A^2
    where x and v are the displacement and velocity at any time, t.
    50*(0.129)^2 + 1*(3.415)^2 = 50*A^2
    50 A^2 = 12.49
    A = 0.500 m
    That does not agreee with your result. sqrt 50 is the angular frequency in Hz, NOT the amplitude A.

    Next write the displacement as
    x = A sin (wt + phi)
    where w is the angular frequency
    w = sqrt (k/m) = 7.071 rad/s
    This and the displacement at t=1 will let you solve for the phase angle phi, and then the amplitude whan t = 0.
    For the velocity at t=0,
    v = dx/dt = w A cos (wt + phi)
    Use the same value of phi.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question