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Period of a cut spring

  1. Dec 20, 2006 #1
    1. The problem statement, all variables and given/known data

    A simple harmonic oscillator consists of a mass m and an ideal spring with spring constant k. Particle oscillates as shown in (i) with period T. If the spring is cut in half and used with the same particle, as shown in (ii), the period will be :
    (choose one of the following)
    T
    [tex]\sqrt{2}T[/tex]
    2T
    T/[tex]\sqrt{2}[/tex]
    T/2


    2. Relevant equations

    [tex]k=m\omega^2[/tex]
    ??

    3. The attempt at a solution

    I blush, but I admit it: guess and check (there were five choices, and we have 5 chances to submit each problem...):redface:

    I figure since this is an ideal spring, it has no mass, so the only equation I can think of is rather useless...
    Any pointers on the right way to figure this baby out?
     

    Attached Files:

  2. jcsd
  3. Dec 20, 2006 #2

    Hootenanny

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    What is the expression for the time-period of an ideal mass-spring system? (You must know this equation :tongue2: it's a manipulation of the formula you quoted). In addition, I should point out that the m in your expression is not the mass of the spring itself, rather the mass hanging from the spring.
     
    Last edited: Dec 20, 2006
  4. Dec 20, 2006 #3
    oh ghod... excuse me while I bury myself in a hole.
    Cramming is evil. It fries brain cells at a time when every brain cell counts.
     
  5. Dec 20, 2006 #4

    OlderDan

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    I'm trying to figure out why the question asked by Hootenanny elicited this response from you. What does happen when you cut the spring in half?
     
  6. Dec 20, 2006 #5
    Because I thought I could sove it. But I couldn't.
    I don't know what happens when you cut a spring in half. However, an educated guess I can make.
    A smaller spring will have a smaller amplitude (and so by default a smaller k). Does the amplitude change by the same amount that the spring was cut (ie spring cut in half==>x_m cut in half)?
    I would think not, because with that assumption I got:

    [tex]T=\frac{2\pi}{\omega}[/tex]

    so if omega=omaga/2, then

    [tex]T_{new}= \frac{4\pi}{omega}=2T=WRONG[/tex]
     
  7. Dec 20, 2006 #6

    Hootenanny

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    Are you sure about that? :wink: And the equation I was refering to above is ;

    [tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]
     
    Last edited: Dec 20, 2006
  8. Dec 20, 2006 #7

    OlderDan

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    Here is a hint then: How far does the middle of the spring move when the end originally attached to the mass moves a distance x?
     
  9. Dec 20, 2006 #8
    It can't be x, because the spring isn't rigid. I mean, the end attatched to a ceiling or wall or whatever doesn't move at all! So the end attatched to the mass moves x, and as you move farther and farther from the mass, the spring moves less and less. But I don't know how to quantify it.
     
  10. Dec 20, 2006 #9

    Doc Al

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    Here's another hint: Say you stretch a spring by an amount X. This takes a certain amount of force. Now cut the spring in half. Will it take more or less force to stretch that half spring by the same distance X?
     
  11. Dec 20, 2006 #10

    Chi Meson

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    What Dan was getting at:
    since you can consider a spring made up of two smaller springs attached end to end, and since the first spring will pull on the second with the same tension (same as the overall pull), then compare the distance moved by the spring attached to the wall to the overall distance moved by the two springs in combination (it's half the distance isn't it?). Since the same force will extend each spring half the distance (x/2) of the total extension (x), what then is the spring constant (the F over x ratio) for one of the springs (if "k" is the constant for the two in combination)?

    Edit:

    Sorry OlderDan, I stepped on yr toes
     
  12. Dec 20, 2006 #11
    I'm going to meet with my physics professor now, so hopefully he'll explain it. If not... I'l be ba-aack. :evilgrin:
     
  13. Dec 21, 2006 #12
    One you find the answer, realize that a simple pendulum has a different equation and a different result when you cut the length in half
     
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