Frequency for resonance to occur

[Helly123]

Homework Statement

That is a speaker. Sound wave is sent out from spesker S into pipe of uniform thickness.
Piston P move to left
1st resonance at 0.045 m
2nd resonance at 0.151 m
Frequency of the sounx 1620 Hz

Piston is stopped at the position that 2nd resonance occur
Then, frequency increased litle by litle
Find frequency at which resonance occur again
End correction is 0.008 m
$\lambda$ = 0.21 m
Velocity = 340 m/s

Homework Equations

The Attempt at a Solution

Frequency = Velocity/$\lambda$

L2 = 0.151 m = 3/4 $\lambda$
$\lambda$ = 4/3 (0.151)m =0.2

Frequency = 340 / 0.20 = 1700 Hz

How to find frequecy when resonance occur again?
What will change the frequency, the velocity, or $\lambda$ ?

 

[kuruman]

What will change the frequency, the velocity, or $\lambda$ ?

What do you think? If you move from a larger to a smaller room, does the speed of sound change?

 

[tnich]

Homework Statement

View attachment 225043
That is a speaker. Sound wave is sent out from spesker S into pipe of uniform thickness.
Piston P move to left
1st resonance at 0.045 m
2nd resonance at 0.151 m
Frequency of the sounx 1620 Hz

Piston is stopped at the position that 2nd resonance occur
Then, frequency increased litle by litle
Find frequency at which resonance occur again
End correction is 0.008 m
$\lambda$ = 212 m
Velocity = 340 m/s

Homework Equations

The Attempt at a Solution

Frequency = Velocity/$\lambda$

L2 = 0.151 m = 3/4 $\lambda$
$\lambda$ = 4/3 (0.151)m =

Frequency = 340 / 0.20 = 1700 Hz

How to find frequecy when resonance occur again?
What will change the frequency, the velocity, or $\lambda$ ?

Notice that your calculated speed of sound does not match the speed of sound given in the problem statement. What are you going to do with the end correction?

 

[tnich]

Notice that your calculated speed of sound does not match the speed of sound given in the problem statement. What are you going to do with the end correction?

Oops. I meant to say your calculated frequency does not match the frequency given in the problem statement.

 

[Helly123]

What do you think? If you move from a larger to a smaller room, does the speed of sound change?

It changes. Larger room = faster?

 

[Helly123]

Oops. I meant to say your calculate frequency does not match the frequency given in the problem statement.

I calcuated the frequency when occur again, after stopped.
Frequency of 1620 Hz is when 1st and 2nd resonance occurred at initial state

 

[kuruman]

It changes. Larger room = faster?

The speed of sound in a ideal gas depends on things like the temperature and the mass of the gas. Here you have air which you may assume is an ideal gas at constant temperature and constant humidity. Therefore, you may safely assume that the speed of sound is constant within the piston.

 

[tnich]

I calcuated the frequency when occur again, after stopped.
Frequency of 1620 Hz is when 1st and 2nd resonance occurred at initial state

I am not sure what you mean by that. Did you use the end correction?

 

[Helly123]

The speed of sound in a ideal gas depends on things like the temperature and the mass of the gas. Here you have air which you may assume is an ideal gas at constant temperature and constant humidity. Therefore, you may safely assume that the speed of sound is constant within the piston.

Therefore? What affect the frequency is the wave lenght?

 

[Helly123]

I am not sure what you mean by that. Did you use the end correction?

Ok. I used End correction to find the lambda before.

First. There are 2 resonances occurred at 0.045 m and 0.151 m . With end correction 0.008 m, frequency 1620 Hz, and lambda is 0.21 m.
Then, the piston stopped at 0.151 m

After that, we want to make the sound to resonance again. At what frequency it is?

 

[Helly123]

This is the full version of the question
In case. I gave the question unclear.
And hope the image shows clear too

And number 4
Piston stopped at position which 2nd resonance took place. If frequency of speaker increased little by little at what frequency for resonance to occur again?

 

[tnich]

This is the full version of the question
In case. I gave the question unclear.
And hope the image shows clear too
View attachment 225060
And number 4
Piston stopped at position which 2nd resonance took place. If frequency of speaker increased little by little at what frequency for resonance to occur again?

Thanks. It helps to have those details. As you said in post #23, changing the frequency will change the wavelength. You have figured out that the first resonance is at $L_1(λ) = \frac λ 4 - ε$ (where L_1(λ) is the length of the pipe and ε is the end correction) and the second resonance is at $L_2(λ)= \frac {3λ} 4 - ε$. Given a wavelength λ, at what length $L_3(λ)$ will the third resonance occur?

 

[Helly123]

Thanks. It helps to have those details. As you said in post #23, changing the frequency will change the wavelength. You have figured out that the first resonance is at $L_1(λ) = \frac λ 4 - ε$ (where L_1(λ) is the length of the pipe and ε is the end correction) and the second resonance is at $L_2(λ)= \frac {3λ} 4 - ε$. Given a wavelength λ, at what length $L_3(λ)$ will the third resonance occur?

$L_3(λ)= \frac {5λ} 4 - ε$
Right?

 

[tnich]

$L_3(λ)= \frac {5λ} 4 - ε$
Right?

Yes.

 

[Helly123]

Yes.

Lambda when 1,2,3 resonance still the same?
Then frequency not change?

 

[tnich]

Lambda when 1,2,3 resonance still the same?
Then frequency not change?

You have the pieces you need. You that increasing the frequency will decrease the wavelength. You know the length of the tube, so you can use $L_3(λ)= \frac {5λ} 4 - ε$ to calculate λ. You know $fλ=v$, so you can calculate frequency.

 

[Helly123]

You have the pieces you need. You that increasing the frequency will decrease the wavelength. You know the length of the tube, so you can use $L_3(λ)= \frac {5λ} 4 - ε$ to calculate λ. You know $fλ=v$, so you can calculate frequency.

So, $L_3$ = $L_2$ = 0.151 m ?

 

[tnich]

So, $L_3$ = $L_2$ = 0.151 m ?

$L_3(λ')=L_2(λ)=$ 0.151 m

 

[Helly123]

$L_3(λ')=L_2(λ)=$ 0.151 m

Do you mean that. We want to have 3rd resonance at the length of 0.151. Hence, we increase the frequency?

 

[tnich]

Do you mean that. We want to have 3rd resonance at the length of 0.151. Hence, we increase the frequency?

I think that is just a restatement of the question in your original post:

Piston is stopped at the position that 2nd resonance occur
Then, frequency increased litle by litle
Find frequency at which resonance occur again

The only additional assumption is that frequency at which resonance occurs again is the third resonance.

 

[Helly123]

I think that is just a restatement of the question in your original post:

The only additional assumption is that frequency at which resonance occurs again is the third resonance.

Ok. The frequency is increased to get 3rd resonance? While the piston not moving?

 

[Helly123]

You have the pieces you need. You that increasing the frequency will decrease the wavelength. You know the length of the tube, so you can use $L_3(λ)= \frac {5λ} 4 - ε$ to calculate λ. You know $fλ=v$, so you can calculate frequency.

$L_3(\lambda')$ = $\frac{5}{4}\lambda$ - ε
0.151 = $\frac{5}{4}\lambda$ - 0.008
$\frac{5}{4}\lambda$ = 0.159
$\lambda$ = 0.1272 m
But it is wrong. Why?

 

[tnich]

$L_3(\lambda')$ = $\frac{5}{4}\lambda$ - ε
0.151 = $\frac{5}{4}\lambda$ - 0.008
$\frac{5}{4}\lambda$ = 0.159
$\lambda$ = 0.1272 m
But it is wrong. Why?

Why do you say it is wrong?

 

[Helly123]

Why do you think it is wrong?

Frequency = v/lambda = 340/0.1272 = 2,672 Hz
It's not the answer

 

[tnich]

Frequency = v/lambda = 340/0.1272 = 2,672 Hz
It's not the answer

You have accumulated some round-off errors. I get a nice round number for the frequency.

 

[Helly123]

You have accumulated some round-off errors. I get a nice round number for the frequency.

Should i round the answer to 0.13?
If it rounded to 0.2 m
It will get 1700 Hz nice number

 

[tnich]

Should i round the answer to 0.13?
If it rounded to 0.2 m
It will get 1700 Hz nice number

Nope. How did you get 340 m/s for the speed of sound?

 

[Helly123]

Nope. How did you get 340 m/s for the speed of sound?

Lambda * frequency = 0.21 m * 1620 Hz = 340.2 m/s

 

[tnich]

Lambda * frequency = 0.21 m * 1620 Hz = 340.2 m/s

With no round-offs, I got speed of sound = $\frac 4 3 (.151+.008) 1620 = 343.44$ m/s. That is not a big difference, so maybe that is not the problem, but it does give a round number for the final frequency.

 

[Helly123]

$L_3(\lambda')$ = $\frac{5}{4}\lambda$ - ε
0.151 = $\frac{5}{4}\lambda$ - 0.008
$\frac{5}{4}\lambda$ = 0.159
$\lambda$ = 0.1272 m
But it is wrong. Why?

Maybe
$L_3(\lambda')$ = $\frac{5}{4}\lambda$ - ε
0.151 = $\frac{5}{4}\lambda$ - 0.008
$\frac{5}{4}\lambda$ = 0.159
$\lambda$ = 0.2 * 4/5 = 0.16 = 0.2 m

That way, frequency = 340/0.2 = 1700 Hz

 

[tnich]

Maybe
$L_3(\lambda')$ = $\frac{5}{4}\lambda$ - ε
0.151 = $\frac{5}{4}\lambda$ - 0.008
$\frac{5}{4}\lambda$ = 0.159
$\lambda$ = 0.2 * 4/5 = 0.16 = 0.2 m

That way, frequency = 340/0.2 = 1700 Hz

Here is what I think happened. Whoever made the answer key thought $L_3(λ') = \frac 4 5 λ'-ε$. That would give 1710hz for the frequency. Rounding to 2 significant digits, it would be 1700hz.

 

[Helly123]

Here is what I think happened. Whoever made the answer key thought $L_3(λ') = \frac 4 5 λ'-ε$. That would give 1710hz for the frequency. Rounding to 2 significant digits, it would be 1700hz.

How can you assume the 1710 Hz? How to get that result?

 

[tnich]

How can you assume the 1710 Hz? How to get that result?

Using $\frac 4 5 λ$ instead of $\frac 5 4 λ$, but that would not be correct.

 

[Helly123]

Ok thanks