Finding a frequency to create a given amplitude in a spring


by ViXXoR
Tags: amplitude, frequency, spring
ViXXoR
ViXXoR is offline
#1
Oct31-07, 07:05 PM
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:
[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
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learningphysics
learningphysics is offline
#2
Nov1-07, 01:46 AM
HW Helper
P: 4,125
You made a mistake here:

[tex]\beta o = \sqrt{\frac{6.30}{0.155}} = 40.6452 rad/s[/tex]
You forgot to take the square root.
ViXXoR
ViXXoR is offline
#3
Nov1-07, 10:41 AM
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.

learningphysics
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#4
Nov1-07, 11:11 AM
HW Helper
P: 4,125

Finding a frequency to create a given amplitude in a spring


Quote Quote by ViXXoR View Post
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.
do you know the answer?

I get 4.095Hz. is that what you're getting?
ViXXoR
ViXXoR is offline
#5
Nov1-07, 11:30 AM
P: 6
Yes, that's exactly what I am getting. It is still marked as wrong though.
learningphysics
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#6
Nov1-07, 11:40 AM
HW Helper
P: 4,125
Quote Quote by ViXXoR View Post
Yes, that's exactly what I am getting. It is still marked as wrong though.
And they want the answer in Hz?
ViXXoR
ViXXoR is offline
#7
Nov1-07, 11:43 AM
P: 6
Yes, they want the answer in Hz.

It seems they want 2 frequencies, a "Lower" one and a "Higher" one.
learningphysics
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#8
Nov1-07, 11:47 AM
HW Helper
P: 4,125
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]
ViXXoR
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#9
Nov1-07, 11:48 AM
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...
learningphysics
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#10
Nov1-07, 12:01 PM
HW Helper
P: 4,125
Quote Quote by ViXXoR View Post
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:

[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 ?
ViXXoR
ViXXoR is offline
#11
Nov1-07, 01:28 PM
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?
CaptainKamel
CaptainKamel is offline
#12
Feb2-09, 11:33 PM
P: 1
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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