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A tube is open only at one end. A certain harmonic produced by the tube has a frequency of 475 hz. The next higher harmonic has a frequency of 1425 hz. The speed of sound in air is 343 m/s

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The formula for the fundamental frequency for an open tube at one end is : F = nV/4L the integer n = 3 which describe the frequency 450 hertz and length of tube L = 0.6m approximately

What is frequency?

Frequency of the object is defined as the number of cycles attend in a second .

Given that a tube is open only at one end. And a certain harmonic produced by the tube has a frequency of 450 Hz. The next higher harmonic has a frequency of 750 Hz. The formula for the fundamental frequency for an open tube at one end is :

[tex]F = \dfrac{V}{4L}[/tex]

That is, 1st harmonic

[tex]\dfrac{V}{4L} = F[/tex]

2nd harmonic =

[tex]\dfrac{3V}{4L} = 3F[/tex]

3rd harmonic =

[tex]\dfrac{5V}{4L} = 5F[/tex]

The general formula can therefore be [tex]F =\dfrac{ nV}{4L}[/tex]

For the first harmonic in the question,

[tex]450 = \dfrac{343n}{4L}[/tex]

make 4L the subject of the formula

[tex]4L = \dfrac{343n}{450}[/tex]

The next higher harmonic will be

[tex]750 = \dfrac{343(n + 2)}{4L}[/tex]

make 4L the subject of the formula

[tex]4L = \dfrac{343(n + 2)}{ 750}[/tex]

Equate the two 4L

[tex]\dfrac{343n}{450} = \dfrac{343( n + 2)}{750}[/tex]

cross multiply

[tex]257250n = 154350( n + 2 )[/tex]

[tex]257250n = 154350n + 308700[/tex]

collect the likes term

[tex]257250n - 154350n = 308700[/tex]

[tex]102900n = 308700[/tex]

[tex]n =\dfrac{ 308700}{102900}[/tex]

n = 3

Given that the speed of sound in air is 343 m/s, The length of the tube can be calculate by using the same formula.

[tex]F = \dfrac{nV}{4L}[/tex]

[tex]450 = \dfrac{3(343)}{4L}[/tex]

[tex]4L = \dfrac{1029}{450}[/tex]

4L = 2.28

[tex]L = \dfrac{2.28}{4}[/tex]

L = 0.57 m

To know more about Frequency follow

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