Suppose a low brass instrument, such as a tuba, has a fundamental frequency of 32.5 Hz. Such instruments act like resonance tubes which are closed at one end, although their resonant frequencies are complicated because they are smaller at the mouth piece and larger at the bell end. For this problem, you may assume the instrument is a simple tube.(a) What is its first overtone, in hertz?
(b) What is its second overtone, in hertz?
(c) What is its third overtone, in hertz?

A. 97.5Hz

B. 162.5 Hz

C. 227.5 Hz

Explanation:

The overtone for a pipe with one closed end is given as:

f(n) = (n + 2) * f(0)

Where f(0) = fundamental frequency

n = odd numbers, 1, 3, 5

A. The first overtone, f(1) of the fundamental frequency, f0 = 32.5 Hz, is given as:

f(1) = 3 * 32.5

f(1) = 97.5Hz

B. The second overtone, f(2) of the fundamental frequency, f0 = 32.5 Hz, is given as:

f(2) = (3+2) * 32.5 = 5 * 32.5

f(2) = 162.5Hz

C. The third overtone, f(3) of the fundamental frequency, f0 = 32.5 Hz, is given as:

f(3) = (5 + 2) * 32.5 = 7 * 32.5

f(3) = 227.5Hz