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Spring with mass on horizontal surface

  1. Apr 30, 2014 #1
    1. The problem statement, all variables and given/known data

    A 2.00 kg object on a horizontal frictionless surface is attached to a spring with
    a spring constant 1000 N/m. The object is displaced from equilibrium 50.0 cm
    horizontally and given an initial velocity of 4.0 m/s away from the equilibrium
    position.

    (a) What is the frequency of motion?

    (b) What is the initial potential energy of the block-spring system?

    (c) What is the initial kinetic energy?

    (d) What is the amplitude of oscillation?


    2. Relevant equations

    F = -kx
    F = ma
    [itex]\omega[/itex] = √(k/m)
    T = 1/f
    f = [itex]\omega[/itex]/2[itex]\pi[/itex]
    PE = 0.5kx2
    KE = 0.5mv2

    3. The attempt at a solution

    (a) So I said:

    T = 1/f → T = 2[itex]\pi[/itex]/f → T = 2[itex]\pi[/itex]√(m/k)

    I subbed in my values and got T = 0.28s

    Then I got f from f = 1/T and got f = 3.57s-1

    (b) From PE = 0.5kx2 I got PE = 125J

    (c) From KE = 0.5mv2 I got KE = 16J

    (d) Its the amplitude part here that I got stuck. Could anyone clarify if my methods for the other parts were correct and point me some direction for part (d). Any help is much appreciated.
     
    Last edited: Apr 30, 2014
  2. jcsd
  3. Apr 30, 2014 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    (c) is weird. check.
    for (d) you can use the fact that at displacement from equilibrium = amplitude value, the movement changes direction, so kinetic energy = 0.
     
  4. Apr 30, 2014 #3
    Hey BvU thanks for the help. (c) should be 16 I forgot to square the 4. Ok so I can say 0.5mv2 = 0 but how to I get an 'A' value from this?
     
  5. Apr 30, 2014 #4
    Wait I think I have it now.

    So PE + KE = 0.5kA2

    KE = 0
    PE = 125

    So work to get A from there?

    I got A = 0.5m which is the same as x so I've a feeling this is wrong.
     
    Last edited: Apr 30, 2014
  6. Apr 30, 2014 #5
    So I have another attempt :tongue:

    The work done decreases the kinetic energy to zero so can I say:

    0.5kA2 = 0.5mv2

    Solving for A = 0.179m
     
  7. Apr 30, 2014 #6
    The spring was given an initial velocity + an initial displacement. So, you have factor both of these things if you are using the ##1/2kA^2## equation, remember that this is equal to the total mechanical energy in your system. It would make no sense for your spring to have a smaller amplitude if it would have more energy in the system as opposed to having no initial velocity.
     
  8. Apr 30, 2014 #7
    Ok so how do I factor in the velocity and displacement? I thought I was with the 'v' in 0.5mv2 and the x in 0.5kx2.
     
  9. Apr 30, 2014 #8
    What is the initial energy put into your system? Don't even worry about the amplitude yet, just figure out the total energy in the system based on the initial displacement from equilibrium and the initial velocity.
     
  10. Apr 30, 2014 #9
    E = KE + PE = 125 + 16 = 141J

    Do I let that equal to the work energy equation then?
     
  11. Apr 30, 2014 #10
    If that's your total mechanical energy, and your know that ##1/2kA^2 = E_{total}## , then you have all you need it looks like.
     
  12. Apr 30, 2014 #11
    Yep I have the answer worked out, thanks a million for the help. Studying for my finals and this really speeds up the process when I'm stuck.
     
  13. Apr 30, 2014 #12
    Glad to hear it, good luck on your finals!
     
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