Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: I freaking hate oscilliations

  1. Jan 11, 2005 #1
    A 1.00 kg object attached to a spring of force constant 41.0 N/m oscillates on a horizontal, frictionless track. At t = 0, the object is released from rest at x = -3.00 cm. (That is, the spring is compressed by 3.00 cm.)

    Find the displacement, velocity and acceleration as functions of the time t.

    my answer:

    my legend:
    w= angular frequency
    &= phase constant

    ok, so for displacement: x(t)=Acos(wt+&)
    velocity: -wAsin(wt+&)
    acceleration: -w^2Acos(wt+&)

    A = max displacement, which is .03 m in this case

    w= 2(pi)(f)

    where f=1/T

    where T=2(pi)(square root of m/k)

    so solving for all that, i get w=6.41 which i know is right...

    so now how do i solve the question?? HOw do i solve each of those equations as a function of something? i have no idea how to do this, i been stuck on this problem forever and i keep getting the wrong answer

  2. jcsd
  3. Jan 11, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    1.If the mass and the elestic constant is give,then it's given [itex] \omega [/itex]

    2. Take the general solution for the periodic motion
    [tex] x(t)=A\sin(\omega t+\phi) [/tex]
    and impose the codition that,at the initial time,the x must be "-A".You'll get the phase.Then to get 'v' and 'a' u need to differentiate wrt time.

  4. Jan 11, 2005 #3
    do you mean cos instead of sin??

    also a quick quesiton, when calculating, do i use radian or degree mode in my graphing calc?

  5. Jan 11, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    1.The anwer will come out with "-cos".
    2.I don't know what will work with your computer... :confused:

  6. Jan 11, 2005 #5
    Use radians for your calculator unless you're very sure that you're working with degrees.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook