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Time period of SHM

  1. Sep 11, 2016 #1
    1. The problem statement, all variables and given/known data

    a man sits in a car that makes the center gravity of the car is pulled down by 0.3 cm. After he gets out of the car, find the time period of the car while it is moving in SHM

    Mass of the car = 500kg
    Spring constant = 196,000 N/m

    2. Relevant equations

    3. The attempt at a solution
    I know that mass and the spring constant, both affect the time period of the object but how is their equation?
    Last edited: Sep 11, 2016
  2. jcsd
  3. Sep 11, 2016 #2


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    Staff: Mentor

    The mass and spring constant are given data, not relevant equations.

    Check your course text or notes or simply search the web for the period of a mass-spring oscillator.
  4. Sep 11, 2016 #3
    sorry, I've edited it.
    I found this equation
    T=2π(SQRT*K/M) but I don't know if it correct or not. I plugged the the spring constant and the mass in the equation, I got about 124.33 s, I think it is too high for the time period
  5. Sep 11, 2016 #4


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    Staff: Mentor

    Presumably you meant:
    $$T = 2 \pi \sqrt{\frac{k}{M}}~~~~~~~~\text{(which is not correct)}$$
    You should be able to check an equation to see if the units make sense. Remember that a Newton (N) can be broken down as ##N = kg~m~s^{-2}## (which should be obvious from Newton's second law, f = ma).

    In this case I think you'll find that the k and M should switch positions. Where did you find your version of the equation?
  6. Sep 12, 2016 #5
    I got it from another physic group
  7. Sep 12, 2016 #6


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    Science Advisor
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    Gold Member

    Can you not see the error in the very last step?
  8. Sep 12, 2016 #7
    When he converted ω is equal to 2π/T?
    then it should be

    ω = (SQRT*k/M)
    2π/T = (SQRT*k/M)
    T = 2π / (SQRT*k/M)

    am I right? I used this equation an I got about 0.317 s
  9. Sep 12, 2016 #8


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    Staff: Mentor

    That works. The answer looks good.

    The formula is usually written as ##T = 2 \pi \sqrt{\frac{M}{k}}##.

    In the scanned page that you attached in post #5 there was an error in the final step of the derivation, as pointed out by @haruspex. It resulted in the argument under the square root being the reciprocal of what it should.
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