Frequency of a wave at resonance

In summary: I'm working with decimals.I should observe my calculator better as well.I think this may be a problem with the text or the way the problem is written. I'll have to ask my prof to see if he can help me out.In summary, the tension in the wire is 647.07 Newtons. The frequency of the air column in the tube is 71.45833333 Hertz.
  • #1
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Homework Statement


A 1.2 m tube is closed at one end. A stretched wire is placed near the open end. The wire is
0.330 m long and has a mass of 9.6 g. It is fixed at both ends and oscillates in its fundamental
mode. By resonance, it sets the air column in the tube into oscillation at that column’s
fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

I only have a problem finding the frequency since the wave speed is not given. Because both the wire and the air are oscillating at their fundamental frequency there must be a way to use this to find the frequency without the wave speed.

Homework Equations


kdx/dt -w = 0 => dx/dt = w/k = f λ => v = f λ => f = v/λ

3. The Attempt at a Solution

λstring = 2Lstring
λair = 4Ltube
fstring = fair
vstringstring = vairair => vstring/2Lstring = vair/4Ltube => vair = 7.42vstring
 
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  • #2
Are you sure that you are not given the speed of sound in air?
 
  • #3
Yes. Only the information given is available. Really makes me think that this was actually a mistake and that it cannot be solved. After thinking about it there is no way to cancel out the velocities with the known information making it impossible to find the frequency. Since this is only a year 1 physics question about waves it shouldn't be difficult anyways.

I've tried emailing my prof about this but apparently he ignores emails...
 
  • #4
If this is a question from a textbook, you might check to see if the there is a statement at the beginning of the exercises/problem section that givens the velocity of sound to use.
 
  • #5
This question actually came from my prof. He claims to have used this on a previous exam. I got a .doc copy of his last exam. this is the 2nd question. Also though I did notice that in the .doc file the last 2 question had some numbers over lapping each other making it impossible to read. So it's entirely possible that some of the information is simply missing or in the wrong format. I'll try reading through my textbook again to see if he took the question from it.

It would make sense for the wave speed of the air to be given since you still have to do everything I already tried to get both the frequency and tension.
 
  • #6
OK. I think this is a problem form the Halliday Resnick Walker text.

Your work looks correct to me.
 
  • #7
I found the question: Halliday Resnick 10th edition chapter 17 page 509 Q#50. It's the exact question from the book. No changes to the question have been made so apparently it should be possible to solve. However the book does not have a solution for it. The fact that my Prof used this on an exam makes believe there's a solution were missing.
 
  • #8
At the beginning of the problem section the text says to use a specific velocity for the speed of sound.
 
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  • #9
Well there's my problem right there. Guess i'll post the solution for anyone interested.
vair = 343m/s
λstring = 2Lstring
λair = 4Ltube
fstring = fair = 71.45833333Hz

vstringstring = vairair => vstring/2Lstring = vair/4Ltube =>vstring = 47.1625m/s
u = m/Lstring = 0.29090909kg
vstring = √t/u => t = u√vstring = 0.066N

Thanks for the help :)
 
Last edited:
  • #10
Send BoBs said:
vstring = √t/u => t = u√vstring = 0.2N
Check this.
 
  • #11
TSny said:
Check this.

should be 0.066N. 0.0096kg⋅√47.01625 = 0.066N
 
  • #12
You didn't solve for t correctly here:
vstring = √t/u => t = u√vstring
 
  • #13
Oh, I took the tension out of the equation: f = 1/λ⋅√t/μ which makes μλ2f2 = t = 647.07N

Weird how I think 0.2N makes sense for a tension force on a string but 600N seems too high.
 
  • #14
Send BoBs said:
Oh, I took the tension out of the equation: f = 1/λ⋅√t/μ which makes μλ2f2 = t = 647.07N
Decimal point is not in the correct place. I think the mistake goes back to your value for u.

Weird how I think 0.2N makes sense for a tension force on a string but 600N seems too high.
Good observation.
 
  • #15
TSny said:
Decimal point is not in the correct place. I think the mistake goes back to your value for u.

Good observation.

I should observe my calculator better as well. I may notice that I'm reading 64.707 and not 647.07
 

What is resonance?

Resonance is the phenomenon that occurs when a vibrating object or system is subjected to an external force that matches its own natural frequency, resulting in a dramatic increase in amplitude of the vibration.

What is the frequency of a wave at resonance?

The frequency of a wave at resonance is the same as the natural frequency of the vibrating object or system. This frequency can be calculated using the equation f = 1/T, where f is the frequency and T is the period of the wave.

How does resonance affect the amplitude of a wave?

Resonance causes a significant increase in the amplitude of a wave. This is because the energy transferred from the external force to the vibrating object or system is at its maximum when the frequency of the force matches the natural frequency of the object or system.

What are some real-life examples of resonance?

Some common examples of resonance include the vibrations of a guitar string when plucked, the sound produced by blowing into a bottle, and the oscillations of a swing.

What factors affect the frequency of a wave at resonance?

The frequency of a wave at resonance depends on the physical properties of the vibrating object or system, such as its mass, stiffness, and length. It is also affected by the external force applied and the medium through which the wave travels.

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