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Calculating the maximum velocity & acceleration of a spring mass

  1. Mar 31, 2012 #1
    1. The problem statement, all variables and given/known data

    A mass of 0.3kg is suspended from a spring of stiffness 200Nm. If the mass is displaced by 10mm from it's equilibrium position and released, for the resulting vibration calculate:

    a) the frequency of vibration
    b) the maximum velocity of the mass during the vibration
    c) the maximum acceleration of the mass during the vibration
    d) the mass required to produce double the maximum velocity calculated using the same spring and initial deflection.

    2. Relevant equations

    w=√k/m
    f=w/2Л
    x=Asin(wt+Ø)
    v=Aw cos(wt+Ø)
    a=Aw² sin(wt+Ø)

    3. The attempt at a solution

    I have used w=√k/m to get w=√200/0.3 w=25.81 rads-1

    The used f=w/2Л to get f=25.81/2x3.142 f=4.108 Hz

    So i am ok working out the frequency of the vibration but i do not understand how to use the remaining equations to get the next answers, i'm not following the process. I'm not asking for answers here, just for someone to help me understand what i need to do to use these and achieve my answers.

    Can anyone help me? Thank you in advance for any responses.

    Daniel
     
  2. jcsd
  3. Mar 31, 2012 #2
    [tex] v= A \omega \cdot cos(\omega t)[/tex]
    A and omega are constants, so v is at it's max when [tex] cos(\omega t)[/tex] is at it's max
     
  4. Apr 1, 2012 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Daniel! :smile:
    I don't understand what you're not getting about this …

    dumbperson :smile: is correct ……

    you have the equations for v and a …

    what is the difficulty? :confused:
     
  5. Apr 1, 2012 #4
    Thanks for the reply guys...

    So, taking v=Aw cos(wt+Ø)

    I get v = 0.01 x 25.81 cos(25.81t+Ø)
    v= 0.2581 cos(25.81t+Ø)

    So multiplying Aw (0.01 x 25.81) gets me 0.2581.

    What happens with the second part of the equation cos(25.81t+Ø)?

    That's leaving me a little confused.
     
  6. Apr 1, 2012 #5
    The cos(and sin) of an angle can vary from a minimum value of zero to a maximum value of one.
    It follows that v is a maximum when cos(wt+ phi) has its maximum value.
    Similar reasoning can be used to find a max.
     
  7. Apr 2, 2012 #6
    Thanks, so is 0.2581 mm/s-1 my final answer for vmax or do i need to work out cos(wt+ phi). If so how do i work out t and phi?
     
  8. Apr 2, 2012 #7
    There is no need to do anything else with the angle other than take the maximum value of cos as being equal to one.In other words vmax=Aw.I haven't checked your numbers but you have presented your units incorrectly.
     
  9. Apr 2, 2012 #8
    Many thanks for your patience :) Your help is much appreciated!
     
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