Final answer:
One of the jet engines has an individual frequency of 4100.250 Hz, and the other engine has an individual frequency of 4099.750 Hz, calculated using the average and beat frequencies.
Step-by-step explanation:
The student asks about the individual frequencies of two jet engines on an airplane that produce an average sound frequency of 4100 Hz with a beat frequency of 0.500 Hz. The individual frequencies of the engines can be found using the concept of beat frequency, which is the absolute difference between the frequencies of two waves.
To find the individual frequencies, we can denote one engine's frequency as f1 and the other's as f2. Given the average frequency (fa) is 4100 Hz, and the beat frequency (fb) is 0.500 Hz, we use the formulas:
- fa = (f1 + f2) / 2
- fb = |f1 - f2|
Using these two equations, we can solve for f1 and f2:
- f1 = fa + fb/2
- f2 = fa - fb/2
Therefore, the individual frequencies of the engines are:
f1 = 4100 Hz + 0.250 Hz = 4100.250 Hz
f2 = 4100 Hz - 0.250 Hz = 4099.750 Hz
One engine is emitting sound at 4100.250 Hz, and the other at 4099.750 Hz.