1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple harmonic frequency clarification

  1. Apr 2, 2007 #1
    A massless spring with spring constant 23.5 N/m hangs vertically. A body of mass 0.30 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released.

    What is the frequency of oscillation of the resulting motion, assumed to be simple harmonic?

    ok so f=(1/2pi)[sqroot(k/m)] cuz w=sqroot(k/m) and f=w/2pi

    so i have (1/2pi)[sqroot(23.5/.30)] which is 13.90.....right?

    and the units would be Hz right becuase w is in rad/s and that divided by rad would give cycles/s


    i think i am doing something wrong. please point it out to me.
     
  2. jcsd
  3. Apr 2, 2007 #2

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    Your thinking and your equation are OK. Maybe your calculator is broken :smile:

    Hint: my calculator says sqrt(23.5/0.3) = 8.85
     
  4. Apr 2, 2007 #3
    thats what i get too but the formula says that 8.85 should be divided by 2pi which is 13.90

    is that calculation right? should i do that?
     
  5. Apr 2, 2007 #4

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    On my calculator 8.85 x PI / 2 = 13.90. I guess you are pushing the wrong buttons.

    2pi is about 6.28. There is no way that 8.85/6.28 could be 13.90 !!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?