Posted by Sarah on Sunday, March 30, 2008 at 9:37pm.
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  drwls, Monday, March 31, 2008 at 11:26am
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.
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