# Finding a frequency to create a given amplitude in a spring

by ViXXoR
Tags: amplitude, frequency, spring
 P: 6 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: $$m=0.155kg$$ $$k=6.30N/m$$ $$F=1.70N$$ $$A=0.440m$$ 2. Relevant equations So a given equation is: $$A = \frac{\frac{F}{m}}{\sqrt{\beta^2 - \beta o^2 + (\frac{b\beta}{m})^2}}$$ Also: $$\beta o = \sqrt{\frac{k}{m}}$$ so $$\beta o = \sqrt{\frac{6.30}{0.155}} = 40.6452 rad/s$$ And: $$f = \frac{\beta}{2\pi}$$ 3. The attempt at a solution Damping is negligible so $$(\frac{b\beta}{m})^2}} = 0$$ Rearranging the first equation for $$\beta$$: $$\beta = \sqrt{\frac{(\frac{F}{m})^2}{A^2} + \beta o^2$$ Plug in all the values: $$\beta = \sqrt{\frac{(\frac{1.70}{0.155})^2}{0.440^2} + 40.6452^2} = 47.6799 rad/s$$ Now using the formula for frequency: $$f = \frac{\beta}{2\pi} = \frac{47.6799}{2\pi} = 7.5885 Hz$$ It seems my answer is wrong, and I cannot find out where I am going wrong. Any advice would be greatly appreciated. Thanks
HW Helper
P: 4,125

 $$\beta o = \sqrt{\frac{6.30}{0.155}} = 40.6452 rad/s$$
You forgot to take the square root.
 P: 6 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.
HW Helper
P: 4,125

## Finding a frequency to create a given amplitude in a spring

 Quote by ViXXoR 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.

I get 4.095Hz. is that what you're getting?
 P: 6 Yes, that's exactly what I am getting. It is still marked as wrong though.
HW Helper
P: 4,125
 Quote by ViXXoR Yes, that's exactly what I am getting. It is still marked as wrong though.
And they want the answer in Hz?
 P: 6 Yes, they want the answer in Hz. It seems they want 2 frequencies, a "Lower" one and a "Higher" one.
 HW Helper P: 4,125 Ah... I think I see the problem now... the square root shouldn't be there: $$A = \frac{\frac{F}{m}}{\beta^2 - \beta o^2}$$
 P: 6 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...
HW Helper
P: 4,125
 Quote by ViXXoR 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...
Are you sure?

The formula I'm seeing online is:

$$A = \frac{\frac{F}{m}}{\sqrt{(\beta^2 - \beta o^2)^2 + (\frac{b\beta}{m})^2}}$$

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:

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

$$\beta = +/- 8.0976 rad/s$$

Frequency = +/- 1.289Hz ?
 P: 6 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?
 P: 1 If High frequency is: f_high = ($$\beta$$)/(2*pi) In order to find the low frequency: f_low = ($$\beta$$)/(4*pi) I got mine right that way.

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