A student must determine the speed of sound in air by using the fundamental frequency of a standing wave. The student places a tuning fork in front of the open end of a pipe that is closed at the other end. The closed end of the pipe can be adjusted so that the length of the pipe can be changed. How should the student adjust the closed end of the pipe to determine the first standing wave that can be established in the pipe?

Respuesta :

Answer:

To determine the first standing wave that can be established in the pipe, the student should adjust the tuning fork such that the closed end is L₀ = λ/4 distance from the opened end, after finding the length of the wavelength of the pipe from Lₙ₊₁ - Lₙ = λ/2.

Explanation:

For a pipe with one end closed, the struck tuning fork vibrates, generating sound waves which travel down to the closed end of the pipe and then reflected back to the opened end. As a wave is generated, another wave is reflected such that both waves form a standing wave in the pipe, with a node at the closed end and an anti-node at the opened end.

Therefore the first harmonics is obtained when the length of the pipe is λ/4. That is N → A

L₀ =  λ/4

Where:

N = Node and

A = Anti-node

λ = Wavelength

L = Length of pipe section from opened to closes end

The next higher harmonics consists of a node at the closed end, an anti-node at the opened end, including a half wavelength in between. That is N A N A

Note that the length N →A → N = λ/2

Therefore the length of the pipe section for the second harmonics = λ/2 + λ/4 = 3/4λ

L₁ = 3/4λ

The length of the nth harmonics is given by

Lₙ = [tex]\frac{1}{4}[/tex]·(2·n+1)·λ

For the student to find the required adjustment for the first harmonic, the student should find the wavelength as follows.

The student should find the length of the pipe section of two successive harmonics as follows

Lₙ = [tex]\frac{1}{4}[/tex]·(2·n+1)·λ and Lₙ₊₁ = [tex]\frac{1}{4}[/tex]·(2·(n+1)+1)·λ

The student should find the difference in length of the pipe section for the two harmonics as

Lₙ₊₁ - Lₙ =  [tex]\frac{1}{4}[/tex]·(2·(n+1)+1)·λ - [tex]\frac{1}{4}[/tex]·(2·n+1)·λ = n·λ/2 + λ/2 + λ/4 - (n·λ/2 + λ/4)

Lₙ₊₁ - Lₙ = λ/2

Then, with the known wavelength, the student should adjust the length of the pipe to obtain a pipe section with length L = λ/4.

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