Answer: the correct answer is a) the interger is n= 3
b)the length of the tube L= 0.57 m
Explanation:
For a tube that is open at one end, nth harmonic frequency is given by
fn=n(v/4L)
the next higher harmonic frequency is given by
fn+2=n+2(v/4L)
Now we have the ratio
fn+2/fn = (n+2)/n= 1+2/n
2/n= [(fn+2)/fn]-1 or
n= 2/{[(fn+2)/fn]-1}
(a) In the problem we have
fn = 450Hz and fn+2= 750Hz So,
n=2/[(750/450)-1] =3
then the frequency of 450Hz corresponds to n=3
b) then we have
450=3v/4L so, L= 3v/(4*450) (3*343)/(4*450) = 0.57m
The formula for the fundamental frequency for an open tube at one end is :
F = nV/4L
n = 3 and L = 0.6m approximately
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 = V/4L
That is, 1st harmonic = V/4L = F
2nd harmonic = 3V/4L = 3F
3rd harmonic = 5V/4L = 5F
The general formula can therefore be F = nV/4L
For the first harmonic in the question,
450 = 343n/4L
make 4L the subject of the formula
4L = 343n/450
The next higher harmonic will be
750 = 343(n + 2)/4L
make 4L the subject of the formula
4L = 343(n + 2)/ 750
Equate the two 4L
343n/450 = 343( n + 2)/750
cross multiply
257250n = 154350( n + 2 )
257250n = 154350n + 308700
collect the likes term
257250n - 154350n = 308700
102900n = 308700
n = 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 = nV/4L
450 = 3(343)/4L
4L = 1029/450
4L = 2.28
L = 2.28/4
L = 0.57 m
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