## simple harmonic motion problem, finding velocity as a function of position

1. The problem statement, all variables and given/known data

I need help finding velocity as a function of position, the t in the argument of tan, causes a problem for me when integrating. can someone help, maybe my approach is completly off, i am trying to fing the unknown phi, knowing the velocity and acceletation when position is 0.

matt

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 Recognitions: Homework Help So you have x= Acos(ωt+φ), you should know that when x=0,t=0 meaning that you can get the value of φ easily. As for the integration, once you get φ, it would be easier to start with a=-ω2x v(dv/dx)=-ω2x v dv = -ω2x dx ∫v dv = -ω2∫ x dx.

 Quote by rock.freak667 So you have x= Acos(ωt+φ), you should know that when x=0,t=0 meaning that you can get the value of φ easily. As for the integration, once you get φ, it would be easier to start with a=-ω2x v(dv/dx)=-ω2x v dv = -ω2x dx ∫v dv = -ω2∫ x dx.
the problem doesnt specify that when x = 0, t =0, how can i make such an assumption?

Recognitions:
Homework Help

## simple harmonic motion problem, finding velocity as a function of position

 Quote by Matt1234 the problem doesnt specify that when x = 0, t =0, how can i make such an assumption?
Yes you are right, I thought your expression for x had the sine term in it, sorry about that.

 no problem, i think this boils down to integrating that above statement, im sorry for the poor image quality, my scanner broke. somehow i need phi, im not quite sure how to obtain it. i think parts a and b we intentional to get me thinking in the way of integration.
 Recognitions: Homework Help Science Advisor Matt1234: You can see, at t = t1 = 0, x1 = 0.10 m. In other words, you need to fully deflect the mass, then release it, in order for it to vibrate. Therefore, at t1 = 0, we have, x1 = 0.10 m = (0.10 m)*cos(4*t1 + phi) cos(4*t1 + phi) = 1 4*t1 + phi = acos(1) phi = acos(1) - 4*t1 phi = 0 - 4*0 phi = 0 radTherefore, we have, x(t) = A*cos(omega*t + phi) x(t) = A*cos(omega*t + 0) x(t) = A*cos(omega*t)Likewise, v(t) = -A*omega*sin(omega*t). (c) Let x(t2) = 0.06 m. Therefore, x(t2) = A*cos(omega*t2) 0.06 = 0.10*cos(4*t2)Hint 1: Can you solve for t2? After that, can you compute v(t2)? (d) Hint 2: Can you compute a(t2)? (e) Hint 3: Let x(t3) = 0 m. Can you solve for t3? Hint 4: Let x(t4) = -0.08 m. Can you solve for t4? Try again. Also, please do not post wide images directly to the forum page. Just post a text link to wide images.