an oscillator with period 2.3 ms passes through equilibrium at t = 7.4 ms with v
ID: 1290803 • Letter: A
Question
an oscillator with period 2.3 ms passes through equilibrium at t = 7.4 ms with velocity v = -2.4 m/s. the equation of the oscillator's motion is x(t) = (A) cm * cos ((omega/s)t + (phi)), where A, omega, and phi are unknowns and need to be solved for based on the information provided. I'm stuck on how to get phi. That said, we know the period and since omega*T=2?, omega is equal to 2?/T. For this case, the 2700 (this is rounded, but it's correct) is correct. Also, if we recall that v = omega*A. Solving for A, you get A = v/omega. In this case you have velocity based of m/s and the answer is in cm so you need to convert fom m to cm to get the answer of .0889cm or 0.000889m. As I mentioned previously, I'm at a loss as to how to calculate phi in this instance.Explanation / Answer
w = 2*pi/T = (2*3.14)/(2.3e-3) = 2730.43 rad/s
at t= 7.4 ms
v = A*w
A = V/w = -(2.4)/(2730.43) = 8.789*10^-4 m = 0.087898 cm
at t= 7.4 ms it at equlibrium so x = 0
0 = 0.087898*cos((2730.43*7.4*0.001(+ phi ))
cos((2730.43*7.4*0.001)+ phi )) = 0
2730.43*7.4*0.001)+ phi = pi/2 = 3.14/2 = 1.57
20.205182 + phi = 1.57
phi = -18.63 radians
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.