Question

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

Answer 1

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 :

$F = \dfrac{V}{4L}$

That is, 1st harmonic

$\dfrac{V}{4L} = F$

2nd harmonic =

$\dfrac{3V}{4L} = 3F$

3rd harmonic =

$\dfrac{5V}{4L} = 5F$

The general formula can therefore be $F =\dfrac{ nV}{4L}$

For the first harmonic in the question,

$450 = \dfrac{343n}{4L}$

make 4L the subject of the formula

$4L = \dfrac{343n}{450}$

The next higher harmonic will be

$750 = \dfrac{343(n + 2)}{4L}$

make 4L the subject of the formula

$4L = \dfrac{343(n + 2)}{ 750}$

Equate the two 4L

$\dfrac{343n}{450} = \dfrac{343( n + 2)}{750}$

cross multiply

$257250n = 154350( n + 2 )$

$257250n = 154350n + 308700$

collect the likes term

$257250n - 154350n = 308700$

$102900n = 308700$

$n =\dfrac{ 308700}{102900}$

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.

$F = \dfrac{nV}{4L}$

$450 = \dfrac{3(343)}{4L}$

$4L = \dfrac{1029}{450}$

4L = 2.28

$L = \dfrac{2.28}{4}$

L = 0.57 m

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