The clarinet is well modeled as a cylindrical pipe that is open at one end and closed at the other. Find the wavelength and frequency of the third normal mode of vibration of a clarinet's air column with effective length of 0.401 m. Take 342 m/s for the speed of sound inside the instrument.

Standing waves are formed when a wave collides with an obstacle and is reflected, the sum of the incident wave and the reflected wave forms a standing wave that does not move in time. There are several types of obstacles:

Fixed. When it collides with a wall a node is formed at this point.

Open. When it collides with a change in density, it is considered a moving obstacle, at this point a belly or antinode is produced.

They indicate that the clarinet is approximated to a tube with one end open and the other closed, therefore at the closed end we have a node and at the open end a antinode, in the adjoint we see a diagram of the resonance, the expression for the resonance should be:

L = [tex]\frac{\lambda}{4}[/tex] 1st harmonic

L = [tex]3 \frac{\lambda}{4}[/tex] 3th harmonic

L = [tex](2n-1) \frac{\lambda}{4}[/tex] General term

where L is the length of the tube, λ the wavelength and n an integer.

Ask for the wavelength of the third harmonic m = 3

λ = [tex]\frac{4L}{3}[/tex]

λ = [tex]\frac{4 \ 0.401}{3}[/tex]

λ = 0.5365 m

The speed of the wave is related to the product of the wavelength and the frequency.

v = λ f

f = [tex]\frac{v}{\lambda}[/tex]

Let's calculate.

f = [tex]\frac{342}{0.5365}[/tex]

f = 639.7 Hz

In conclusion, using the characteristics of standing waves in tubes we can find the results for the third harmonic of the clarinet are:

The wavelength is λ = 0.5365 m

The frequency is: f = 639.7 Hz

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