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What is the correct formula for acceleration for SHM

  1. Feb 20, 2015 #1
    1. The problem statement, all variables and given/known data
    If the mass of a horizontal mass-spring oscillator undergoing SHM is 0.5kg and the force constant is 20N/cm, what is the maximum restoring force of the oscillator? And what is the maximum acceleration? (Time period is 6s and amplitude of oscillation is 12cm)

    2. Relevant equations
    Acceleration (a) = -w^2(x) and a = -k(x/m)

    3. The attempt at a solution
    When I use the given information and plug in the values, the answer is different for each equation. For the first equation, I get an answer of 0.13ms^-2, and for the latter I get 480ms^-2.

    I don't seem to understand why they don't match and which one is correct. because I know we had to equate ma=-kx to derive the equation omega(w) = sqr.rt(k/m).

    Any replies would be tremendously helpful. Thanks.

    PS - This is my first post. :)
     
  2. jcsd
  3. Feb 20, 2015 #2

    Nathanael

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    Homework Helper

    Welcome to PF

    I'm not entirely sure I understand what's confusing you:
    You calculated the acceleration using a=-ω2x and got a different answer than when you used a=-kx/m?

    But like you said, ω=√(k/m) so how can the equations be any different?
     
  4. Feb 20, 2015 #3
    Thanks for the reply.... Actually Yes, I did get two different answers for the different formulae. I think the problem lies with the value to ω. When I use

    ω=√(k/m),
    with k = 20N/cm, and m = 0.5kg,
    I get 6.32 rad s^-1 as the answer.

    Plugging in this value of ω in

    a=-ω^2x gives me 4.80 ms^-2

    But when I use

    ω = 2π/T, with T = 6s,
    I get the value 1.05 rad s^-1.

    Thus, a=-ω^2x now gives me 0.13 ms^-2

    I can't figure out why I get two different values. Shouldn't they match? And also, which one is correct?

    PS - I used F =-k.x to find the maximum restoring force. I got 240 N which seems to be correct. Its the acceleration bit thats throwing me off.
     
    Last edited: Feb 20, 2015
  5. Feb 20, 2015 #4

    gneill

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    Staff: Mentor

    The given spring constant and mass imply a natural period of about a tenth of a second. This is a far cry from the suggested 6 seconds. No mention of damping is made (and it would have to be large indeed), so the given values seem self-contradictory.
     
  6. Feb 21, 2015 #5
    Thanks so much..... I was suspecting something must have been wrong with the question. BTW, when you say the information implies a natural period of a tenth of a second, how do you arrive at that conclusion? I don't think I know of any equations or relationships that helps me calculate that. Can you help? :)
     
  7. Feb 21, 2015 #6

    gneill

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    Staff: Mentor

    Angular frequency ω is related to the period. You should know how ω relates to f, and how f relates to T.
     
  8. Feb 21, 2015 #7
    Oh, ofcourse, silly me... :)... Thanks... I think I've figured it out... Cheers
     
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