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jessiebotello7209
05.05.2020 •
Mathematics
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.
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Ответ:
a. 17.145cm
b. 8.325cm
c. 0.846
d. 2.896
Step-by-step 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
Ответ:
Step-by-step explanation:
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