Question

When a sound source of frequency n is approaching a stationary observer with velocity u than the apparent change in frequency is Δn₁ and when the same source is receding with velocity u from the stationary observer than the apparent change in frequency is Δn₂. Then ______.

Options

• Δn₁ = Δn₂

• Δn₁ < Δn₂

• Δn₁ > Δn₂

• Δn₁ = Δn₂ = 0

Answer 1

When a sound source of frequency n is approaching a stationary observer with velocity u than the apparent change in frequency is Δn₁ and when the same source is receding with velocity u from the stationary observer than the apparent change in frequency is Δn₂. Then Δn₁ > Δn₂.

Explanation:

Δn₁ = `"v"/("v"-"v"_"s") "f"-"f"`

Δn₂ = `"f"-"v"/("v"-"v"_"s") "f"`

Δn₁ = `("v"_"s""f")/("v"-"v"_"s")`

Δn₂ = `("v"_"s""f")/("v"+"v"_"s")`

According to the aforementioned equation, the value of the fraction will be larger if the denominator is smaller and smaller if the denominator is greater. Thus, we can say that Δn₁ > Δn₂.