Question about sound waves

In summary, a tuning fork is used to find various positions of maximum loudness in an air-filled tube of variable length. The difference in tube length between adjacent maxima is 10.0 cm. The wavelength of the sound is 0.20m and the frequency of the tuning fork is 1700 Hz. Closing the right hand end of the tube does not change the 10.0 cm measurement as the distance between resonances is still lambda/2.
  • #1
narutodemonki
7
0

Homework Statement


A tuning fork is set up near the end of an air filled tube(open at both ends) of variable length. By changing the length various position of maximum loudness( resonances) can be found. The difference in tube length between adjacent maxima is found to be 10.0 cm
a) wavelength of air of this sound?
b) frequency of tuning fork?
c) would closing the right hand end of tube change 10.0 cm measurement?

Homework Equations

and attempt
for a) I used equation L=n lambda/2, lambda was found to be 0.20m by assuming n=1... not sure why n =1, but i get the right answer. ( did not say 1st harmonic etc.)b) frequency of tuning fork I used v= lambda *f .. and got f=1700 HZ.

c) I am not sure how to answer C) in the answer section it says: no because lambda/2 distance is still between resonance:confused:
what I know is closing the right end would made the anti node become a node.
 
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  • #2
You get the first maximum at different length, but the difference in tube length between adjacent maxima stays the same 10.0 cm.

ehild
 
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  • #3


I would like to provide a more thorough explanation for each part of this question.

a) To find the wavelength of the sound in the air, we can use the equation L = nλ/2, where L is the length of the tube, n is the harmonic number, and λ is the wavelength. In this case, n=1 because we are assuming the first harmonic, which is the fundamental frequency. This means that the length of the tube is equal to half of the wavelength. Therefore, we can rearrange the equation to solve for λ, which gives us λ = 2L = 2(0.10m) = 0.20m.

b) To find the frequency of the tuning fork, we can use the equation v = λf, where v is the speed of sound in air, λ is the wavelength, and f is the frequency. The speed of sound in air is approximately 343 m/s at room temperature. Using the value of λ that we found in part a (0.20m), we can rearrange the equation to solve for f, which gives us f = v/λ = 343 m/s / 0.20m = 1715 Hz. This is the theoretical frequency of the tuning fork.

c) Closing the right hand end of the tube would not change the 10.0 cm measurement. This is because the distance between adjacent maxima is determined by the wavelength of the sound, which is a property of the sound wave itself. The position of the maxima is not affected by the end of the tube being closed. However, closing the end of the tube would change the effective length of the tube, which would change the frequency of the sound produced. This would result in a different wavelength and therefore a different distance between adjacent maxima.
 

What are sound waves?

Sound waves are a type of mechanical wave that travel through a medium, such as air or water, and cause particles in the medium to vibrate. These vibrations create changes in air pressure, which our ears detect as sound.

How do sound waves travel?

Sound waves travel through a medium in a series of compressions and rarefactions. In other words, as the wave moves forward, it causes particles in the medium to move closer together and then farther apart.

What is the speed of sound?

The speed of sound varies depending on the medium it is traveling through. In dry air at room temperature, sound travels at approximately 343 meters per second. However, it can travel faster or slower in different mediums, such as water or solids.

How does sound intensity affect our perception of sound?

Sound intensity refers to the amount of energy carried by a sound wave. The greater the intensity, the louder the sound will seem to our ears. However, our perception of sound intensity is not linear and can vary based on factors such as distance from the source and the frequency of the sound.

How are sound waves used in everyday life?

Sound waves have many practical applications in our daily lives. They are used in communication technologies, such as cell phones and radios, as well as in medical imaging techniques like ultrasounds. Sound waves are also used for navigational purposes, such as in sonar systems used by ships and submarines.

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