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Simple Harmonic Motion period

  1. Nov 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A 775 g mass is hung on a spring. As a result the spring stretches 20.5 cm. If the object is then pulled an additional 3.0 cm downward and released, what is the period of the resulting oscillation?

    2. Relevant equations
    T = 2pi sqr root(m/k)
    Hooke's Law Fs=kx

    3. The attempt at a solution
    For the period I know you use T = 2pi sqr root(m/k)

    So you must find k using Hooke's Law, which we find force (m*g) and total distance is our number plus the additional 3cm, so to solve for k, its k=F/x, which you then substitute into the original equation, but I don't think my answer is right. I'm assuming my x is wrong? Do you not add the two values?

    20.5cm + 3cm = .235m

    (.775)(9.81) / .235m = 32.34

    T = 2pi sqrroot (.775kg/32.34) = .97
    97 seconds.
  2. jcsd
  3. Nov 18, 2015 #2
    How much force does it take to stretch the spring a distance of 20.5 cm?
  4. Nov 18, 2015 #3
  5. Nov 18, 2015 #4
    How did you come up with that answer?
  6. Nov 18, 2015 #5


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    The mass was pulled down the additional 3 cm. You can't use this distance to figure out the spring constant because you don't know the total force used to stretch the spring.
    What are the units here?

    The mass by itself stretches the spring only 20.5 cm according to the problem statement.
    How does .97 become 97 seconds?
  7. Nov 18, 2015 #6
    brought to rest (lowered carefully, so Fnet = 0)
  8. Nov 18, 2015 #7
    I dont know
  9. Nov 18, 2015 #8
    If it takes no force to stretch a spring, that spring won't make the attached object oscillate.
  10. Nov 18, 2015 #9
    So how do I determine the force?
  11. Nov 18, 2015 #10
    I answered the other questions within
  12. Nov 18, 2015 #11
    It was given in the statement of the problem.
  13. Nov 18, 2015 #12
    I'm lost.
  14. Nov 18, 2015 #13
    By definition, ##k=\displaystyle \frac{F}{x}## where ##F## is the force applied to the spring and ##x## is the distance the spring is stretched by that force.
  15. Nov 18, 2015 #14


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    You know that 775 grams stretches the spring 20.5 cm. You use the weight of a 775-gram mass to find the force stretching the spring.
  16. Nov 18, 2015 #15
    So 9.81*.775? ................
  17. Nov 18, 2015 #16
    So what I would change is just use .03m instead of the .235m?
  18. Nov 18, 2015 #17
    What would be your reasoning for doing that?

    Perhaps you need some time to go back and read the problem again and read through this thread again.
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