Need Help on Difficult Wave Questions

  • Thread starter ultra_mc
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  • #1
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I just tried to post this topic but the site logged me out when I tried to post it. Now I have to do it all over again fun!!!

This is page 1: http://s890.photobucket.com/albums/ac101/ultra_mc/?action=view&current=Image.jpg

So question 1:

Homework Statement


f = ?
x = 5.5m
Δx = 2m
v= 340m/s
d = 4.5m


Homework Equations



v = fλ

PnS1 - PnS2 = (n - 1/2)λ



The Attempt at a Solution



Made a triangle to find PnS1:

5.5^2 + 4.5^2 = c^2
c = 7.11
PnS1 = 7.11m

PnS1 - PnS2 = (n - 1/2)λ
7.11 - 5.5 = (1 - 1/2)λ
1.61
---- = λ
0.5

λ = 3.22

v = fλ
340 = f(3.22)
340
--- = f
3.22

f = 105.59Hz

Which is not one of the answers...

What did I do wrong?
 

Answers and Replies

  • #2
Delta2
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I think the mistake is that you substitute n=1. You know that this equation should hold for some n, not necessarily for n=1 or for every n. Use the data that the second minimum is 2m further and make a second equation in a similar way. With the two equations you should able to calculate n and lamda.
 
  • #3
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I think the mistake is that you substitute n=1. You know that this equation should hold for some n, not necessarily for n=1 or for every n. Use the data that the second minimum is 2m further and make a second equation in a similar way. With the two equations you should able to calculate n and lamda.
I have no idea what equation I would use for that since the equations always use n and wavelength.
 
  • #4
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bump ;_;
 
  • #5
Delta2
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Calculate new c^2=7.5^2+4.5^2, new c-7.5. The difference 1.61-(c-7.5) is the wavelength.
 
  • #6
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Calculate new c^2=7.5^2+4.5^2, new c-7.5. The difference 1.61-(c-7.5) is the wavelength.

I don't think that works because that would mean the wavelength is 5.89m. When used for v = f(lambda) then I get 57.72 Hz which isn't an answer either >_<
 
  • #7
ehild
Homework Helper
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The Attempt at a Solution



Made a triangle to find PnS1:

5.5^2 + 4.5^2 = c^2
c = 7.11
PnS1 = 7.11m

PnS1 - PnS2 = (n - 1/2)λ
7.11 - 5.5 = (1 - 1/2)λ
1.61
---- = λ
0.5

λ = 3.22


What did I do wrong?

Do not take n equal to 1. It is an unknown integer. Use the other minimum 2 m out. For that, n is 1 less.

ehild
 
  • #8
Delta2
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I don't think that works because that would mean the wavelength is 5.89m. When used for v = f(lambda) then I get 57.72 Hz which isn't an answer either >_<

Something u did wrong cause i found a wavelength of about 0.36m which gives 944,5Hz.

c_old-5.5=1,61
c_new-7,5=1,25

1,61-1,25=0,36

If you wonder why this is the wavelength it is because the new equation is c_new-7,5=(n-1-1/2)lamda (as ehild told for the new minimum, n is one less). if u subtract this new equation from the old equation, you get (c_old-5.5)-(c_new-7.5)=lambda.
 
Last edited:

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