Question

A trumpet has a length of 140cm. The first valve lowers the root note by two semitones (f1/f0 = 2-2/12) and the second valve lowers the root note by one semitone (f2/f0 = 2-1/12). A) What is the change in length for each case? (i.e. how long is the tubing for the valve slides). B) Now suppose the player presses both the first and second valve (this has the expected effect of lowering the note by three semitones), how much extra tubing is added to the length? (do not round) C) What is the new frequency ratio? D) if the ratio is modeled as f3/f0 = 2-n/12 determine n.

Answer 1

a. 17.145cm

b. 8.325cm

c. 0.846

d. 2.896

Explanation:

It is known that frequency is inversely proportional to length.

Case 1: The player presses only the first valve.

f₁/fo = 2^{-2/12} = lo/l₁

given that lo= 140cm.

l₁ = 2^{2/12}lo = 157.145

change in length:

l₁ - lo = 157.145 - 140 = 17.145cm

Case 2: The player presses only the second valve.

f₂/fo = 2^{-1/12} = lo/l₂

l₂ = 2^{1/12}lo = 148.325cm

l₂ - lo= 148.325-140 = 8.325cm

Case 3: The player presses both the first and second valve.

Change in length is the sum of change in length when they are individually pressed.

l₃ - lo = 17.145+8.325 = 25.47cm

l₃ = 25.47cm+lo = 165.47cm

f₃/fo = 2^{-n/12} = lo/l₃

f₃/fo = 2^{-n/12} = 140/{165.47 = 0.846

2^{-n/12} = 0.846\Rightarrow n/12 = -log(0.846)/log(2) = 0.2413

n = 0.2413*12 = 2.896