Recent content by ubbaken

  1. U

    How Do You Calculate the Phase Constant in Simple Harmonic Motion?

    but it can't be 2pi/3 rads, cause that isn't the phase, it's the next one equal to -1/2. But I've tried them both. It accepts deg or rad answers.
  2. U

    More Simple Harmonic Oscillations

    [FONT="Times New Roman"]An object in simple harmonic motion oscillates with a period of 4.00 s and an amplitude of 9.08 cm. How long does the object take to move from x=0.00 cm to x=5.07 cm? I set up my eqn like this: 0.0908cos(ωt)=0.0507 cos(ωt)=0.583 ωt=56.1 then with ω=90deg I get...
  3. U

    How Do You Calculate the Phase Constant in Simple Harmonic Motion?

    tried that, though I didn't word it that well there it seems. with that logic, I work through to get: -1/2=cos(φ); which gives me 120deg - which isn't in the correct phase, so continuing along the curve until 240deg gives the right phase, but the answer is coming up as incorrect.
  4. U

    How Do You Calculate the Phase Constant in Simple Harmonic Motion?

    What is the phase constant? Use a cosine function to describe the simple harmonic motion. http://capa.physics.mcmaster.ca/figures/kn/Graph14/kn-pic1416_new.png t1=40.0 s and A=20.0 cm I'm really lost on how to get this done. Using x=Acos(ωt+φ); my approach has been to find when...
  5. U

    2-D kinematics/Forces (weight up an incline)

    ah yes, I see now. I was making some incorrect equalities with my angles. Thank you for your help.
  6. U

    2-D kinematics/Forces (weight up an incline)

    That was my first post, I believe.
  7. U

    2-D kinematics/Forces (weight up an incline)

    Sorry, my x' was 'parallel' to the ramp and y' was perpendicular. What I have are: F(normal)=wcosθ and F(push)=wsinθ=(123kg)(9.81m/s/s)sin(26.9deg)=546N But that is incorrect...
  8. U

    2-D kinematics/Forces (weight up an incline)

    The Question: A 123 kg chesterfield is pushed up a frictionless ramp at a constant speed by a delivery person. If the ramp is inclined at 26.9° to the horizontal, what horizontal force must the delivery person apply to the chesterfield...
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