A tube is open only at one end. A certain harmonic produced by the tube has a frequency of 450 Hz. The next higher harmonic has a frequency of 750 Hz. The speed of sound in air is 343 m/s.

(a) What is the integer n that describes the harmonic whose frequency is 450 Hz?

(b) What is the length of the tube?

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