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

1. Oct 31, 2007

### ViXXoR

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

Last edited: Oct 31, 2007
2. Nov 1, 2007

### learningphysics

You forgot to take the square root.

3. Nov 1, 2007

### 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.

4. Nov 1, 2007

### learningphysics

I get 4.095Hz. is that what you're getting?

5. Nov 1, 2007

### ViXXoR

Yes, that's exactly what I am getting. It is still marked as wrong though.

6. Nov 1, 2007

### learningphysics

And they want the answer in Hz?

7. Nov 1, 2007

### ViXXoR

Yes, they want the answer in Hz.

It seems they want 2 frequencies, a "Lower" one and a "Higher" one.

8. Nov 1, 2007

### learningphysics

Ah... I think I see the problem now... the square root shouldn't be there:
$$A = \frac{\frac{F}{m}}{\beta^2 - \beta o^2}$$

9. Nov 1, 2007

### 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...

Last edited: Nov 1, 2007
10. Nov 1, 2007

### learningphysics

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 ?

Last edited: Nov 1, 2007
11. Nov 1, 2007

### ViXXoR

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?

12. Feb 2, 2009

### CaptainKamel

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.