Question

Suppose a low brass instrument, such as a tuba, has a fundamental frequency of 305 Hz. Such instruments act like s:lusu04az.theexpertta.com/Common/ViewAssignmentDetails.aspx 9/2018 The Expert TA l Human-like Grading, Automated! 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.

Part (a) What is its first overtone, in hertz? Numeric :A numeric value is expected and not an expression.
Part (b) What is its second overtone, in hertz? Numeric Aeric value is expected and not an expression.
Part (c) What is its third overtone, in hertz? Numeric :A numeric value is expected and not an expression.

Answer 1

A. 915 Hz

B. 1525 Hz

C. 2135 Hz

Step-by-step explanation:

The overtone of a tube open at one end and closed at the other is given as

fn = (n + 2)f0

Where n is a odd number (1, 3, 5...)

f0 is the fundamental frequency.

A. The first overtone is:

f1 = (1 + 2) * 305 = 3 * 305

f1 = 915 Hz

B. The second overtone is:

f2 = (3 + 2) * 305 = 5 * 305

f2 = 1525 Hz

C. The third overtone is:

f3 = (5 + 2) * 305 = 7 * 305

f3 = 2135 Hz