I came across a physics problem for homework that I was able to work out on my own, however when I got to class, my teacher said that my numbers were not what the book had calculated, but couldn't tell if there was a flaw in my logic that would prevent me from getting the correct phi and therefore the correct answers. The question: A simple harmonic oscillator consists of a block of mass 4.20 kg attached to a spring of spring constant 200 N/m. When t = 4.00 s, the position and velocity of the block are x = 0.129 m and v = 3.415 m/s. My approach a.) What is the amplitude of the oscillations? I used conservation of mechanical energy to solve this part of the problem, and got the correct amplitude this way: (1/2)kx^2 + (1/2)mv^2 = (1/2)k(amplitude)^2 amplitude = .5114 m b.) What was the position of the mass at t = 0 s? For this part, I plugged the position I was given and the time I was given as well as the amplitude and omega, which I solved for, into this formula: x(t) = amplitude * cos(omega * t + phi) I solved for omega: omega = (k/m)^(1/2) omega = 6.9 rad/s and used the amplitude from part a) and the given x and t to solve for phi .129 = .5114 * cos(6.9 * 4 + phi) .252 = cos(27.6 + phi) 1.316 = 27.6 + phi phi = -26.28 rad then I use this phi in the x(t) function to solve for x at t = 0: x(0) = .5114 * cos(-26.28) x(0) = .21 m c.) What was the velocity of the mass at t = 0 s? I use the same phi, amp, and omega from before, except in the equation: v(t) = -omega * amplitude * sin(omega * t + phi) v(0) = -(6.9) * (.5114) * sin(-26.28) v(0) = 3.217 m/s When my teacher does the problem another way, however, he got a phi that was different from mine. Teacher's Approach v(4)/x(4) = [-omega * amplitude * sin(omega * t + phi)] / [amplitude * cos(omega * t + phi)] v(4)/x(4) = -omega * tan(omega * t + phi) 26.473 = -6.9 * tan(6.9 * 4 + phi) -3.837 = tan(27.6 + phi) -1.316 = 27.6 + phi phi = -28.916 rad He then proceeded to solve for the amplitude using x(t) = amplitude * cos(omega * t + phi) and x(0) and v(0) in pretty much the same way that I did. My question is, why did using his method get the equation: -1.316 = 27.6 + phi phi = -28.916 rad when my way instead gave me: 1.316 = 27.6 + phi phi = -26.28 rad ? In order to get the phi that my teacher got, I would have to alter the formula that I used to solve for phi to: x(t) = amplitude * cos(-omega * t - phi) which I don't think is correct. I would greatly appreciate any insight into this matter.