The motors that drive airplane propellers are, in some cases, tuned by using beats. The whirring motor produces a sound wave having the same frequency as the propeller. Part A
If one single-bladed propeller is turning at 574 rpm and you hear a 2.50 Hz beat when you run the second propeller, what are the two possible frequencies (in rpm) of the second propeller? f1,f2 =
Part B
Suppose you increase the speed of the second propeller slightly and find that the beat frequency changes to 2.60 Hz . In part (A), which of the two answers was the correct one for the frequency of the second single-bladed propeller ?
Part C
How do you know the answer in part (B) to be correct? Explain.

A) 2 possible frequencies of second propellar = 424 rpm or 724 rpm

B) Correct frequency is f2 = 724 rpm

C) Reason is stated in explanation

Explanation:

A) We are given;

Frequency of first propeller; f1 = 574 rpm

Beat frequency; f_beat = 2.5 Hz = 2.5 × 60 rpm = 150 rpm

Now, formula for the beat frequency is;

f_beat = |f1 - f2|

Now, |f1 - f2| means it is inside an absolute value.

Thus, it means,

f1 - f2 = 150 or f2 - f1 = 150

Thus;

574 - f2 = 150

Or f2 - 574 = 150

So,f2 = 574 - 150 = 424 rpm or f2 = 574 + 150 = 724 rpm

2 possible frequencies of second propellar = 424 rpm or 724 rpm

B) Now, if we increase the speed of the second propellar slightly, it means that f2 will increase as well.

Now, from the 2 possible values of f2 gotten, we can see that for f1 - f2 = 150, when we increase f2, the beat frequency will reduce while for f2 - f1 = 150, when we increase f2, the beat frequency will increase.

Thus, it's the frequency of f2 gotten in f2 - f1 = 150 that is the correct answer.

Thus, when we increase the speed of the second propellar slightly, it means that;

f2 = 724 rpm

C) Answer in part B is correct because as we increase the speed of the second propellar, the frequency will also increase and the value of f2 that corresponds with an increase in speed is 724 rpm.

none, its a machine