Find Tension for Desired Harmonic Resonance

[Hockeystar]

Homework Statement

You have designed a new musical instrument of very simple construction. Your design consists of a metal tube with length L and diameter L/10. You have stretched a string of mass per unit length u across the open end of the tube. The other end of the tube is closed. To produce the musical effect you're looking for, you want the frequency of the third-harmonic standing wave on the string to be the same as the fundamental frequency for sound waves in the air column in the tube. The speed of sound waves in this air column is v.

EDIT: What must be the tension of the string to produce the desired effect?

Homework Equations

f = nv/4L (n is 1,2,3, ... ect.)

v= (T/u)^0.5

The Attempt at a Solution

f= v/4L = 3/2L*(T/u)^0.5

f= v²*u/36

This answer is wrong I`m sort of lost. Help would be appreciated.

 

[Andrew Mason]

The Attempt at a Solution

f= v/4L = 3/2L*(T/u)^0.5

f= v²*u/36

This answer is wrong I`m sort of lost. Help would be appreciated.

First, you will have to tell us what the question is. What is it you are trying to find? The tension in the string or the frequency?

Second, you will have to explain your reasoning.

It seems to me that you are confusing two speeds. There is the speed of sound in air and there is also the speed of the wave on the vibrating string. They are two very different things.

The frequency is rather simple to determine. You correctly state that the fundamental frequency in the closed tube is f = v/4L where v is the speed of sound in air. That frequency is fixed by L and has nothing to do with the vibrating string. The frequency of the vibrating string, depends on the tension. This tension has to be set so that the third harmonic (how many half wavelengths are there in the length of the string for the third harmonic?) matches the frequency of the air in the tube. What is that tension? Is that not what the question is asking?

AM

 

[Hockeystar]

Sorry, I forgot to include, "What must be the tension of the string to produce the desired effect?". Thanks for correcting me on (v) sound vs. (v) string. I still need some help though.

f = v_sound/4L (Air Column closed Fundemental)
f = 3/2L*(T/u)^0.5 (String third harmonic)

Equate the freq together:

v_sound/4L = 3/2L*(T/u)^0.5
T= (v_sound² )*u/36

This answer is still wrong however. My answer, "contains an incorrect numerical multiplier or is missing one."

 

[Andrew Mason]

...
v_sound/4L = 3/2L*(T/u)^0.5
T= (v_sound² )*u/36

This answer is still wrong however. My answer, "contains an incorrect numerical multiplier or is missing one."

How do you get f = 3/2L*(T/u)^0.5? Explain your reasoning. What are you using for the string length and how many half wavelengths do you have for the third harmonic?

AM

 

[Hockeystar]

Crap, it's a closed air column not an open one. Setting the frequencies together I should get:

T = v²u/9L⁴

 

[Andrew Mason]

Crap, it's a closed air column not an open one. Setting the frequencies together I should get:

T = v²u/9L⁴

Your comment is difficult to understand. Your answer can be easily seen to be wrong because the dimensions are wrong. Tension has dimensions of force or mass x distance x time^-2. You answer has dimensions of mass x distance^-3 x time^-2 The tube is closed on one end. So with a standing wave in that tube, you have one node. So tube length corresponds to 1/4 of a wavelength, as you said initially. See: http://hyperphysics.phy-astr.gsu.edu/hbase/waves/clocol.html#c1

Begin with the universal wave equation: [itex]\lambda f = v[/itex]

For the closed column [itex]\lambda/4 = L[/itex] so [itex]f = v_{sound}/4L[/itex] which is what you had.

For the string, do the same thing:

[tex]f = v_{string}/\lambda_{string}[/tex]

Substitute your equation for speed of the string and then equate the two frequencies. (hint: what is the length of the string? - read the question again and answer my last post).

AM

 

[Hockeystar]

EDIT: Just edited it as you replied. The length should be 1/10L not L. The third harmonic for a closed tube is:

1/10L = 3/4[tex]\lambda[/tex]
[tex]\lambda[/tex]= 4L/30f= v_string/[tex]\lambda[/tex]
f= (T/u)^0.5 *1/(4L/30)
f= (T/u)^0.5 *(30/4L)

Equate the two together:

v_sound/4L = (T/u)^0.5 *(30/4L)
T= v²u/900

 

[Andrew Mason]

Ok I think I finally got it. The third harmonic for a closed tube is:

L = 3/4[tex]\lambda[/tex]
[tex]\lambda[/tex]= 4/3L

f= v_string/[tex]\lambda[/tex]
f= (T/u)^0.5 *1/(4L/3)
f= (T/u)^0.5 *(3/4L)

Equate the two together:

v_sound/4L = (T/u)^0.5 *(3/4L)
T= v²u/9

NO!

Read the question carefully. You do not need to know the third harmonic for the closed tube. You need to know the third harmonic for the vibrating string. There are two things that are vibrating here. The air column and the string. The wavelengths of the vibrating air in the tube and the vibrating string are NOT the same.

WHAT IS THE LENGTH OF THE STRING? That is the key to finding the wavelength of the standing wave on the vibrating string and, therefore, the frequency at which the string vibrates.

AM

 

[Andrew Mason]

EDIT: Just edited it as you replied. The length should be 1/10L not L. The third harmonic for a closed tube is:

1/10L = 3/4[tex]\lambda[/tex]
[tex]\lambda[/tex]= 4L/30

You are not reading the question. The question asks you to find the tension in the string when the frequency of the third harmonic of the string is equal to FUNDAMENTAL frequency of the tube.

The fundamental vibration in the tube occurs when [itex]L = \lambda_{air}/4[/itex]. You got this right the first time.

The third harmonic of the string occurs when the length of the string - which you correctly note is L/10 - is equal to HOW MANY HALF WAVELENGTHS of the standing wave on the string? Call that number n.

[tex]L/10 = n\lambda_{string}/2[/tex]

Use those values for the wavelengths of the air and string. Use the universal wave equation then to determine the frequency at those wavelengths.

AM

 

[Hockeystar]

Thanks for helping me with this problem, it was train wreck for me. Even though I got the first question wrong I got the next 2 right. Finally understood that the string behaves like an open column because there are nodes on each end.

 

[Andrew Mason]

Thanks for helping me with this problem, it was train wreck for me. Even though I got the first question wrong I got the next 2 right. Finally understood that the string behaves like an open column because there are nodes on each end.

No. The string has nodes on each end but an open column has anti-nodes on each end. The standing waves on the string, therefore, must have integral numbers of half wavelengths. The third harmonic occurs with 3 half wavelengths in the standing wave.

For the string:

[tex]3\lambda_{string}/2 = L/10[/tex]

[tex]f_{string} = v_{string}/\lambda_{string} = v_{string}/(2L/30) = \left(\frac{T}{u}\right)^{.5}15/L[/tex]

You previously found [itex]f_{column} = v_{air}/4L[/itex]

Then you just equate the two frequencies and find the Tension.

AM