How Do You Solve Complex Wave Equations Involving Frequency and Wavelength?

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In summary, the conversation is about a physics problem involving finding the frequency of a wave using given values for velocity, distance, and wavelength. The person made a mistake by substituting n=1 in their equations and is advised to use the data from the second minimum to calculate n and lambda. Another person suggests using the equations c_new-7.5=(n-1-1/2)lambda and (c_old-5.5)-(c_new-7.5)=lambda to find the correct wavelength, which is approximately 0.36m and gives a frequency of 944.5Hz.
  • #1
ultra_mc
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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?
 
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  • #2
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
Delta² said:
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
bump ;_;
 
  • #5
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
Delta² said:
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
ultra_mc said:

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
ultra_mc said:
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:

Related to How Do You Solve Complex Wave Equations Involving Frequency and Wavelength?

1. What are some tips for solving difficult wave questions?

Some tips for solving difficult wave questions include: breaking the problem down into smaller parts, using diagrams and sketches, understanding the basic principles and equations related to waves, and practicing with different types of wave problems.

2. How do I identify the type of wave in a question?

To identify the type of wave in a question, you can look for keywords such as "transverse," "longitudinal," or "mechanical." You can also use the wave equation, which includes variables like wavelength, frequency, and amplitude, to determine the type of wave.

3. What is the difference between a standing wave and a traveling wave?

A standing wave is a wave that appears to be stationary and does not transfer energy, while a traveling wave is a wave that transfers energy through a medium. Standing waves are formed by the interference of two traveling waves with the same frequency and amplitude, while traveling waves continue to move through a medium.

4. How can I calculate the speed of a wave?

The speed of a wave can be calculated by dividing the wavelength by the period, or by multiplying the wavelength by the frequency. The equation for wave speed is v = λ/T = λf, where v is the wave speed, λ is the wavelength, and f is the frequency.

5. What are some common applications of wave phenomena?

Wave phenomena are used in various fields, including communication, medicine, and technology. Some common applications include radio and television broadcasting, ultrasound imaging, and seismic exploration for oil and gas. Waves are also used in musical instruments, such as guitars and pianos, to produce sound.

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