
#1
Oct3107, 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 



#2
Nov107, 01:46 AM

HW Helper
P: 4,125

You made a mistake here:




#3
Nov107, 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. 



#4
Nov107, 11:11 AM

HW Helper
P: 4,125

Finding a frequency to create a given amplitude in a springI get 4.095Hz. is that what you're getting? 



#5
Nov107, 11:30 AM

P: 6

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




#6
Nov107, 11:40 AM

HW Helper
P: 4,125





#7
Nov107, 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. 



#8
Nov107, 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] 



#9
Nov107, 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...




#10
Nov107, 12:01 PM

HW Helper
P: 4,125

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 ? 



#11
Nov107, 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?




#12
Feb209, 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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