# Frequency for resonance to occur

• Helly123
In summary: L_1(λ) is the length of the pipe and ε is the end correction) and the second resonance is at ##L_2(λ)=...##. Given a wavelength λ, at what length ##L_3(λ)## will the third resonance occur?If you increase the frequency by little by little, you will get to the frequency where the second resonance occurs again. This is at ##L_3(λ) = \frac {3λ} 4 - ε##.
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

## 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## ?

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• 20180502_224608.jpg
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Last edited:
Helly123 said:
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?

Helly123 said:

## 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

## 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 said:
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.

kuruman said:
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?

tnich said:
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

Helly123 said:
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.

Helly123
Helly123 said:
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?

kuruman said:
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?

tnich said:
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?

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?

#### Attachments

• 20180503_074006.jpg
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Helly123 said:
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?

tnich said:
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?

Helly123 said:
##L_3(λ)= \frac {5λ} 4 - ε##
Right?
Yes.

tnich said:
Yes.
Lambda when 1,2,3 resonance still the same?
Then frequency not change?

Helly123 said:
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.

tnich said:
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 ?

Helly123 said:
So, ##L_3## = ##L_2## = 0.151 m ?
##L_3(λ')=L_2(λ)=## 0.151 m

Helly123
tnich said:
##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?

Helly123 said:
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:
Helly123 said:
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.

tnich said:
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?

Last edited:
tnich said:
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?

Helly123 said:
##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?

tnich said:
Why do you think it is wrong?
Frequency = v/lambda = 340/0.1272 = 2,672 Hz

Helly123 said:
Frequency = v/lambda = 340/0.1272 = 2,672 Hz
You have accumulated some round-off errors. I get a nice round number for the frequency.

tnich said:
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

Helly123 said:
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?

tnich said:
Nope. How did you get 340 m/s for the speed of sound?
Lambda * frequency = 0.21 m * 1620 Hz = 340.2 m/s

Helly123 said:
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 said:
##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

Helly123 said:
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.

tnich said:
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?

Helly123 said:
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.

Ok thanks

## 1. What is resonance frequency?

Resonance frequency is the natural frequency at which an object vibrates with the greatest amplitude when excited by an external force.

## 2. How is resonance frequency calculated?

Resonance frequency can be calculated using the formula f = 1/2π√(k/m), where f is the resonance frequency, k is the object's stiffness, and m is its mass.

## 3. Why is resonance frequency important?

Resonance frequency is important because it can cause an object to vibrate with a large amplitude, which can lead to damage or failure if not controlled properly. It is also used in various technologies such as musical instruments and radio receivers.

## 4. How does resonance frequency affect sound?

In sound, resonance frequency refers to the frequency at which an object or medium vibrates most efficiently, resulting in a louder sound. This is why musical instruments have certain resonance frequencies that produce distinct tones.

## 5. How can resonance frequency be controlled?

Resonance frequency can be controlled by changing the stiffness or mass of an object, or by using damping techniques to reduce the amplitude of vibrations. In some cases, adding a counterweight or changing the shape of an object can also alter its resonance frequency.

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