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.

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

first, we define the variables:

x: number of years after 1950

f (x): amount of vinyl sold.

then, with the variables defined, we have:

68594 vinyl records were sold in 1958 > f (8) = 68594

91299 vinyl records were sold in 1961 > f (11) = 91299

38720 vinyl records were sold in 1952 > f (2) = 38720

161743 vinyl records were sold in 1967 > f (17) = 161743

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step-by-step explanation:

good luck on your studies! : )

david edward.