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!

Finding a frequency to create a given amplitude in a spring

  1. Oct 31, 2007 #1
    1. The problem statement, all variables and given/known data
    Damping is negligible for a 0.155 kg object hanging from a light 6.30 N/m spring. A sinusoidal force with an amplitutde of 1.70 N drives the system. At what frequency will the force make the object vibrate with an amplitude of 0.440 m?
    So:
    [tex]m=0.155kg[/tex]

    [tex]k=6.30N/m[/tex]

    [tex]F=1.70N[/tex]

    [tex]A=0.440m[/tex]


    2. Relevant equations
    So a given equation is:

    [tex]A = \frac{\frac{F}{m}}{\sqrt{\beta^2 - \beta o^2 + (\frac{b\beta}{m})^2}}[/tex]

    Also:

    [tex]\beta o = \sqrt{\frac{k}{m}}[/tex] so [tex]\beta o = \sqrt{\frac{6.30}{0.155}} = 40.6452 rad/s[/tex]

    And:

    [tex]f = \frac{\beta}{2\pi}[/tex]


    3. The attempt at a solution
    Damping is negligible so [tex](\frac{b\beta}{m})^2}} = 0[/tex]
    Rearranging the first equation for [tex]\beta[/tex]:

    [tex]\beta = \sqrt{\frac{(\frac{F}{m})^2}{A^2} + \beta o^2[/tex]

    Plug in all the values:

    [tex]\beta = \sqrt{\frac{(\frac{1.70}{0.155})^2}{0.440^2} + 40.6452^2}

    = 47.6799 rad/s[/tex]

    Now using the formula for frequency:

    [tex]f = \frac{\beta}{2\pi}

    = \frac{47.6799}{2\pi}

    = 7.5885 Hz[/tex]


    It seems my answer is wrong, and I cannot find out where I am going wrong. Any advice would be greatly appreciated.

    Thanks
     
    Last edited: Oct 31, 2007
  2. jcsd
  3. Nov 1, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    You made a mistake here:

    You forgot to take the square root.
     
  4. Nov 1, 2007 #3
    Wow...

    Thank you for pointing that out.

    Hmmmmm......turns out that when I take the square root, the answer is more wrong. I'm really stumped.
     
  5. Nov 1, 2007 #4

    learningphysics

    User Avatar
    Homework Helper

    do you know the answer?

    I get 4.095Hz. is that what you're getting?
     
  6. Nov 1, 2007 #5
    Yes, that's exactly what I am getting. It is still marked as wrong though.
     
  7. Nov 1, 2007 #6

    learningphysics

    User Avatar
    Homework Helper

    And they want the answer in Hz?
     
  8. Nov 1, 2007 #7
    Yes, they want the answer in Hz.

    It seems they want 2 frequencies, a "Lower" one and a "Higher" one.
     
  9. Nov 1, 2007 #8

    learningphysics

    User Avatar
    Homework Helper

    Ah... I think I see the problem now... the square root shouldn't be there:
    [tex]A = \frac{\frac{F}{m}}{\beta^2 - \beta o^2}[/tex]
     
  10. Nov 1, 2007 #9
    If I use that equation, how will I get 2 frequencies out of it? Also, the square root is in the equation in my textbook...
     
    Last edited: Nov 1, 2007
  11. Nov 1, 2007 #10

    learningphysics

    User Avatar
    Homework Helper

    Are you sure?

    The formula I'm seeing online is:

    [tex]A = \frac{\frac{F}{m}}{\sqrt{(\beta^2 - \beta o^2)^2 + (\frac{b\beta}{m})^2}}[/tex]

    which isn't what you had... you didn't have the extra square inside... when there's no damping... the square root and square cancel each other.

    Maybe the 2 values are the plus/minus square root...

    So:

    [tex]\beta = +/-\sqrt{\frac{(\frac{F}{m})}{A} + \beta o^2[/tex]

    [tex]\beta = +/- 8.0976 rad/s[/tex]

    Frequency = +/- 1.289Hz ?
     
    Last edited: Nov 1, 2007
  12. Nov 1, 2007 #11
    Awesome, you're right! One thing, can you have a negative frequency? Can I have my higher frequency at 1.289hz, and my lower one at -1.289Hz?
     
  13. Feb 2, 2009 #12
    If High frequency is:

    f_high = ([tex]\beta[/tex])/(2*pi)

    In order to find the low frequency:

    f_low = ([tex]\beta[/tex])/(4*pi)


    I got mine right that way.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Finding a frequency to create a given amplitude in a spring
Loading...