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!

Harmonic oscillators homework

  1. Nov 30, 2008 #1
    Two different simple harmonic oscillators have the same natural frequency (f=3.40 Hz) when they are on the surface of the Earth. The first oscillator is a pendulum, the second is a vertical spring and mass. If both systems are moved to the surface of the moon (g=1.67 m/s^2, what is the new frequency of the pendulum? Calculate the new frequency of the vertical spring and mass.


    Any idea on the problem... i dont have the minimum idea to solve it....
     
  2. jcsd
  3. Nov 30, 2008 #2
    Re: Frequency....Help

    So find out how gravity affects frequency, and compare the earth's gravity (appx. 9.78m/s^2) to the moon's (1.67m/s^2).
    Oh, and you'll also need to know how an oscillator and a vertical spring and mass work, respectively.
     
  4. Nov 30, 2008 #3
  5. Nov 30, 2008 #4
    Re: Frequency....Help

    Oh.. didn't see the "simple harmonic" part.. well thanks for replying as well Bright Wang, you seem to be more help than me :P
     
  6. Nov 30, 2008 #5
    Re: Frequency....Help

    w=2*pi*f

    but w=sqrt(k/m) and k=(mg)/x_displacement

    so f=sqrt(g/x_disp)*[1/(2*pi)]

    and x_disp= ( g / [(f)*(2*pi)]^2 )
    x_disp= (9.8/ [(3.40)*(2*pi)]^2)
    = 2.15x10^-2

    if i use now the gravity of the moon it is

    f = sqrt(g/x_disp)*[1/(2*pi)]
    = sqrt(1.67/2.15X10^-2)*[1/(2*pi)]
    = 1.40

    and this is the wrong answer for frequency of the vertical sring and mass...

    can anyone help me...
     
  7. Nov 30, 2008 #6
    Re: Frequency....Help

    the frequncy 1.40 Hz is the answer for the pendulum but in my calculus it is sopose to be the same for both sistems and that is wrong... any suggestion ....
     
  8. Nov 30, 2008 #7
    Re: Frequency....Help

    1) for the pendulum f=(1/(2pi))[g/L]^1/2 so f~(g)^1/2, and I got 1.40 Hz

    2) for the spring T=1/f=2(pi)[M/k]^1/2 , so its independent?
     
  9. Nov 30, 2008 #8
    Re: Frequency....Help

    ok but if i calculate the frquency for the spring it is the same.... and that is not the correct answer...
     
  10. Nov 30, 2008 #9
    Re: Frequency....Help

    what is the correct answer?
     
  11. Nov 30, 2008 #10
    Re: Frequency....Help

    ok i understand the gravity just affect the pendulum not the spring... ty
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Harmonic oscillators homework
  1. Harmonic oscillator (Replies: 9)

  2. Harmonic Oscillator (Replies: 15)

  3. Harmonic Oscillations (Replies: 10)

  4. Harmonic oscillator (Replies: 2)

  5. Harmonic Oscilator (Replies: 5)

Loading...