Answer:
960.24 Hz
Explanation:
Here is the complete question
A sound wave of the form
S=Smcos(kx−ωt+Φ)
travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?
Solution
Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm
Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.
So, s = s₀cos(kx−ωt+Φ)
6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1
cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0
Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.
So, s = s₀cos(kx−ωt+Φ)
2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3
cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.
We now have two simultaneous equations.
2k + Φ = 0 (1)
2.070k + Φ = 70.53. (2)
Subtracting (2) -(1)
2.070k - 2k = 70.53
0.070k = 70.53
k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m
k = 2π/λ ⇒ λ = 2π/k
and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz
