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Mass suspended on a spring question

  1. Sep 4, 2007 #1
    1.5kg mass on spring streatched 3cm from natural length. oscillationg amplitude of 2.2cm. when t=0.01s, displacement x = 1.5 cm, i'm assuming gravity of 9.81 m/s

    I've worked out k to be 490.5 N, w to be 18.08, T to be 0.348 secs and f to be 2.97Hz, which i think are all correct.

    the question asks find displacement x as a function of time, I dont even know what this means, I dont even know what formula to use.
  2. jcsd
  3. Sep 4, 2007 #2


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    To find displacement as a function of time, you find an expression describing the system with the displacement variable [tex] x[/tex] and the time variable [tex] t[/tex], and then you re-arrange the expession to isolate the displacement variable on one side of the equation and an expression for time on the other side, i.e., find [tex] x(t) [/tex].
  4. Sep 4, 2007 #3
    Yeah that makes sense and was kinda the area I was thinking.
    I've just found this:

    x(t) = xm(the amplitude of motion) cos 2*pi*t/T

    and think this might be the one to use, not sure how this will work out though
  5. Sep 4, 2007 #4
    i've now got x(t) = 0.0632 does this seem right? and is it the final answer? i'm slightly worried as i have not used the x=1.5cm????
  6. Sep 4, 2007 #5


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    Homework Helper

    The general equation is:

    [tex]x(t) = Acos(\omega{t} + \phi) + x_0[/tex]

    where [tex]x_0[/tex] is your equilbrium position... this is the position where the mass hangs from the spring without oscillating... ie: x0 = mg/k... a position of x = 0, corresponds to the spring's unstretched position.

    you already know A and [tex]\omega[/tex]... you need to get [tex]\phi[/tex]

    Edit: switched from y to x.
    Last edited: Sep 4, 2007
  7. Sep 4, 2007 #6
    So the position i am at now is:- (I hope my earlier calculations arn't wrong)

    x(t) = 0.022 cos(18.08 * 0.01 + (pi / 2)) + 0.015 ?

    so my answer is:
    x(t) = 0.037 ( does this even sound reasonable?)

    My course notes and the books I have are of no help at all on this question. I don't even know how to work out the phase angle(phi), the best i can come up with is phi = pi/2 ?? which seems too simple

    I'm now so confused and my head hurts. It doesn't help that my assessor is away for 2 weeks either.

    Any extra help would be greatly recieved.
    Last edited: Sep 4, 2007
  8. Sep 5, 2007 #7


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    All your calculations for the first parts look right. The second part is asking for displacement as a function of time... that means t will be a variable in your expression...

    [tex]x(t) = Acos(\omega{t} + \phi) + x_0[/tex]

    substituting your A, omega, and x0(x0 is actually given in the question as 3cm)

    [tex]x(t) = 0.022cos(18.08{t} + \phi) + 0.03[/tex]

    once you get phi, just replace [tex]\phi[/tex] with the number... and then you're done.

    To solve for [tex]\phi[/tex] use the fact that at t = 0.01s, x = 0.015m

    So plug in t = 0.01, and x(t) = 0.015 and solve for [tex]\phi[/tex] in radians.
  9. Sep 5, 2007 #8

    0.015 * 0.01 = 0.022cos(18.08t + phi) + 0.03

    0.00015 / 0.022 = cos(18.08 * 0.01 + phi) + 0.03

    0.0068 - 0.03 = cos(0.1808 + phi)

    -0.0232 = cos0.1808 + cos phi

    -0.0232 / cos1 = cos phi

    -0.0232 = cos phi

    cos -1(-0.0232) = phi

    phi = 91.33 rad/s ????????? I'm sure i've got this totally wrong
  10. Sep 5, 2007 #9


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    By the x(t)... I didn't meant x multiplied by t... but just meant to emphasize that x is a function of t... you can just leave it as x if you want.

    [tex]x = 0.022cos(18.08{t} + \phi) + 0.03[/tex]

    So plug in x = 0.015 and t = 0.01

    0.015 = 0.022cos(18.08*0.01 + phi) + 0.03

    0.015 -0.03 = 0.022cos(0.1808 + phi)

    -0.015 = 0.022cos(0.1808 + phi)
    -0.015/0.022 = cos(0.1808 + phi)

    Now take the inverse cosine of both sides (use your calculator for the left side)... then solve for phi...
  11. Sep 6, 2007 #10
    I've just spoke to another instructor at the uni and he gave me the formula

    X = A*sin(wt+phi)

    I solved for phi and got 40.58 rad/s

    Does this look right anyone?

    Now I assume i put phi into the formula with A and w and leave X and t as letters?
  12. Sep 6, 2007 #11


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    You can use that formula as long as you define the x = 0 position as being when the mass is hanging motionless from the string... ie the equilibrium position...

    It really depends on what they mean by displacement x... do they mean displacement from the unstretched spring position... or do they mean displacement from the 3cm position...

    Do they give a picture by any chance? Anyway, phi should between 0 and 2pi = 6.28radians...

    Reading the question again, yeah, they're probably asking for displacement from the position where the mass is hanging at rest... so

    X = A*sin(wt+phi)

    should be fine.

    Can you show your calculations for how you got phi?
    Last edited: Sep 6, 2007
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