One of the harmonics of a column of air in a tube that is open at one end and closed at the other has a frequency of 448 Hz, and the next higher harmonic has a frequency of 576 Hz. What is the fundamental frequency of the air column in this tube?
a. 32 Hz.
b. 64 Hz.
c. 128 Hz.
d. 256 Hz.
e. 88 Hz.

The correc answer is c. 128 Hz

Explanation:

The lowest resonant frequency of a vibrating object is called the fundamental frequency. Most vibrating objects have more than one resonance frequency. The frequency of the nth mode is n times the frequency of the fundamental mode: fn = n · fo

You know that a column of air in a tube that is open at one end and closed at the other has a frequency of 448 Hz, and the next higher harmonic has a frequency of 576 Hz. So you know that

fn=448 Hz=n*fo (A)

and in the next harmonic fn+1=576 Hz= (n+1)*fo (B)

You can make the difference between (B) and (A):

(B)-(A)

576 Hz - 448 Hz= (n+1)fo - nfo

128 Hz=(n+1-n)fo

128 Hz=fo

So the difference between the frequencies of the two consecutive harmonics is the fundamental frequency of the air column in the tube.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west