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Springs - Expressions

  1. Nov 23, 2009 #1
    1. A mass attached to the lower end of a vertical spring causes the spring to extend by 25 mm to its equilibrium position. The mass is then displaced a further 20 mm and released. A trace of the vibration and time measurements are taken. From these measurements it can be seen that the displacement from the equilibrium position is 19.2 mm when the time is 0.05 s.)
    (a) Write the expression for the displacement of the mass as a function of time.
    (b) Write the expression for the velocity of the mass as a function of time.
    (c) Write the expression for the acceleration of the mass as a function of time.




    2. a. Y= ASinWT, b. v= AwCoswT, c. a= -Aw^2Sin wT



    3. I have calculated x = 0.0192, t = 0.05, w = 19.8, A = 0.02

    Attempt

    a. Y = 20Sin19.8t
    b. v = 396Cos19.8t
    c. a= -7840.8Sin19.8t

    Any advice on where I am or what I am doing wrong appreciated, thanks
     
  2. jcsd
  3. Nov 23, 2009 #2

    Doc Al

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    Staff: Mentor

    Since the mass is released at its maximum displacement from equilibrium, you'd be better off using cosine instead of sine to describe the displacement.

    How did you compute ω?
     
  4. Nov 23, 2009 #3
    W = sqrt K/M = sqrt 9810/25 = W^2 = 9810/25 = 19.8

    I am actually quite lost as to what I have to achieve
     
  5. Nov 23, 2009 #4

    Doc Al

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    Staff: Mentor

    Good.
    I presume the position and time were given so you can calculate the phase at t = 0. (I guess it isn't at maximum displacement at t = 0, as I had first thought. A bit strangely worded, I'd say.)

    Use a more general form for the displacement; something like y = A sin(ωt + Φ).
     
  6. Nov 23, 2009 #5
    Thanks, Do I insert values ie: A = 0.02xSin

    So phase angle is 0
     
  7. Nov 23, 2009 #6
    Then work out a figure for each epression e V = ****
     
  8. Nov 23, 2009 #7

    Doc Al

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    No. You know A and ω. Insert the given values for y and t, then you can solve for the phase.
     
  9. Nov 24, 2009 #8
    How does this look:

    y = A sin(wt + b) - you are given x = 0.0192 when t = 0.05, A = 0.02 and you know w, so you can find a
    d) y = A sin (wt + b)
    A = 20 mm, w = 19.8 rad/s
    Now y = 19.2 mm at t=0.05s
    Substituting
    19.2 = 20 Sin(19.8x0.05 + b)
    Thus, Sin(0.99 + b) = 19.2/20
    Sin(0.99 + b) = 0.96
    0.99 + b = Sin-10.96 = 1.287 ( Take the angle in radians)
    So b = 1.287 – 0.99 = 0.297
    Or y = 20 Sin(19.8t + 0.297)

    e) Similarly, v =396 Cos(19.8t + 0.297)

    f) a = --7840.8 Sin(19.8t + 0.297)
     
  10. Aug 28, 2010 #9
    (a) Write the expression for the displacement of the mass as a function of time.

    x = Asin (wt + 0)

    (b) Write the expression for the velocity of the mass as a function of time.

    V = dx/dt = Aw cos (wt + 0)


    (c) Write the expression for the acceleration of the mass as a function of time.

    a = dv/dt = -Aw2 sin (wt + 0)
     
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