How Does Mass Affect the Oscillation Frequency of a Spider's Web?

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The discussion centers on calculating the effective spring stiffness constant of a spider's web and predicting the oscillation frequency when a heavier insect is trapped. The initial calculation for the stiffness constant k was confirmed as 0.22 N/m. However, participants faced challenges in determining the new frequency for a 0.50g insect, with initial answers exceeding the original frequency of 4.4 Hz, which is incorrect since increased mass should lower the frequency. A mistake in calculator usage was identified as a common issue among participants, leading to confusion in arriving at the correct frequency. Ultimately, the correct approach emphasizes that the frequency must decrease with increased mass, aligning with the principles of oscillation physics.
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Homework Statement



A small fly of mass 0.29g is caught in a spider's web. The web oscillates predominately with a frequency of 4.4Hz .

a) What is the value of the effective spring stiffness constant k for the web?

b) At what frequency would you expect the web to oscillate if an insect of mass 0.50g were trapped?

Homework Equations



f = (1/2Pi)(k/m)^.5

The Attempt at a Solution



a) rearranging the equation

k =m(2Pif)^2 = (.00029kg)(2Pi*4.4Hz)^2 = .22 N/m

mastering physics says I am right on this one. So I figured for part b I just rearrange the equation for f and plug in .0005 for m and .22 for k

f=(1/2Pi)(k/m)^.5 = (1/2Pi)(.22N/m/.0005kg)^.5 = 33Hz

but mastering physics says i am wrong. Where is my mistake?
 
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The expression is correct. An estimate (sqrt(2000/5) = 20) yields 20/π, around 3.
As expected: twice as heavy, 1.4 times lower freq.
 
Hmm... I'm not sure, looks like right to me. using w = root(k/m) and w = 2pif i get the same thing. Did you try 32.9hz?
 
it said 32.9 was wrong also. and it says in the question it wants 2 sig figs =[
 
Have you tried adding another sig fig? use .22 for k and .0005 for m and take it out 2 places.
 
without rounding the answer is 32.94930172. I am not sure if that's what you mean
 
Yea that's about what I got, but it wants 2 sig figs and 33 isn't working, so maybe they mis stated and want 2 after the decimal? I'm honestly not sure, toothpaste. Your algebra looks good, we got the same answer, and assuming it did in fact say that .22N/m was the correct k value (I didn't double check that), I'd probably just punch it in the face and call it a day, honestly.

Btw, I've been meaning to say something for a while, but I think your name is effing hilarious XD
 
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yeah its telling me .22 N/m is correct on the first part. =[ so frustrating.

and thank you! :D i love this username!

anyways thanks for your help i guess ill email my professor, maybe there is a problem with mastering physics.
 
Listen up, you two: if a.29 g fly swings with 4.5 Hz a HEAVIER fly swings at a LOWER freq. So anything above 4.4 Hz is plain wrong.
 
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  • #10
By increasing the mass the frequency must decrease. You should be aware that 30 something does not make sense if the original frequency was 4.4 Hz.

Check your units and calculations.
 
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  • #11
(1/2Pi)(.22N/m/.0005kg)^.5 is correct.

You must be doing something wrong with your calculator. You are getting the decimal point in the wrong place.
 
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  • #12
Yea I see what you guys are saying, I kind of feel like a noob for not seeing that. That's my bad everyone =[

I just think it's weird we both got the same (wrong) thing.
 
  • #13
I forgot parenthesis on my calculator. I entered 20.976/2Pi instead of 20.976/(2Pi)... sorry guys. I got 3.3 this time and mastering physics accepted it
 
  • #14
My point is that at all times you should make a reasonable guess about the outcome:
1. The square root can't be far from 20,
2. The freq has to be lower than 4.4 by a factor of around 1.5
 
  • #15
Yea i should have noticed that. Sometimes the intuition aspect of it doesn't come very easily to me
 
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