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12 July, 09:11

A sound wave of the form s = sm cos (kx - ? t + f) 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?

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  1. 12 July, 11:57
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    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
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