Final answer:
The displacement of an element of air located at x = 2.4 m at time t = 6.7 ms is 9.89 µm.
Step-by-step explanation:
The displacement of an element of air located at x = 2.4 m at time t = 6.7 ms can be calculated using the given wave function s = sm cos (kx - ωt).
First, we need to find the values for k and ω using the given information. The wave number k can be obtained from the formula k = 2π/λ. Thus, k = 2π/0.55m = 11.42 m^(-1). The angular frequency ω can be calculated using the formula ω = 2πf, where f is the frequency. Therefore, ω = 2π * 384 Hz = 2413.44 Hz.
Now, we can substitute these values into the wave function to find the displacement of the element of air at x = 2.4 m and t = 6.7 ms:
s = sm cos((11.42 m^(-1) * 2.4 m) - (2413.44 Hz * 6.7 ms))
s = 14 µm * cos(27.41 - 16.14) = 14 µm * cos(11.27) = 9.89 µm.