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
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
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