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Homework Help: 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?




    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]


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


    [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.

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


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    Homework Helper

    You made a mistake here:

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

    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


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    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


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    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


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    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


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    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...


    [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.
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