Doppler effect — all four cases for sound, formula selection

Question

What are the four cases of the Doppler effect for sound, and how do we select the correct formula for each?

Solution — Step by Step

The general Doppler effect formula for sound:

f' = f_0 \left(\frac{v \pm v_o}{v \mp v_s}\right)

where:

$f'$ = apparent frequency heard by observer
$f_0$ = actual frequency of source
$v$ = speed of sound in the medium
$v_o$ = speed of observer
$v_s$ = speed of source

Sign convention: Take the direction from source to observer as positive.

Case 1: Source moves toward stationary observer

f' = f_0 \left(\frac{v}{v - v_s}\right) > f_0

Frequency increases — pitch goes up.

Case 2: Source moves away from stationary observer

f' = f_0 \left(\frac{v}{v + v_s}\right) < f_0

Frequency decreases — pitch goes down.

Case 3: Observer moves toward stationary source

f' = f_0 \left(\frac{v + v_o}{v}\right) > f_0

Frequency increases.

Case 4: Observer moves away from stationary source

f' = f_0 \left(\frac{v - v_o}{v}\right) < f_0

Frequency decreases.

Use the SOS rule (Source Opposite Sign):

Observer approaching → add $v_o$ in numerator (towards = increase = add)
Source approaching → subtract $v_s$ in denominator (towards = source opposite sign = subtract)

Or simply remember: any approach increases frequency, any recession decreases it. Then check: does the formula give $f' > f_0$ for approach? If not, flip the sign.

When both source and observer move:

f' = f_0 \left(\frac{v + v_o}{v - v_s}\right) \text{ (both approaching)}

f' = f_0 \left(\frac{v - v_o}{v + v_s}\right) \text{ (both receding)}

If source approaches but observer recedes (or vice versa), mix the signs accordingly.

flowchart TD
    A["Doppler Effect Problem"] --> B{"Who is moving?"}
    B -->|"Source only"| C{"Direction?"}
    B -->|"Observer only"| D{"Direction?"}
    B -->|"Both"| E["Use general formula with appropriate signs"]
    C -->|"Toward observer"| F["f prime = f0 times v over v minus vs"]
    C -->|"Away from observer"| G["f prime = f0 times v over v plus vs"]
    D -->|"Toward source"| H["f prime = f0 times v plus vo over v"]
    D -->|"Away from source"| I["f prime = f0 times v minus vo over v"]

Why This Works

When a source moves toward you, the wavelength gets compressed (source catches up with its own waves). Shorter wavelength at the same wave speed means higher frequency. When you move toward the source, you encounter wave crests more frequently — so frequency appears higher.

The key physics: source motion changes the wavelength in the medium, while observer motion changes the rate of encountering wavefronts. Both affect the perceived frequency, but through different mechanisms.

Alternative Method

For quick numerical checks, use the percentage approach. If $v_s \ll v$ , the fractional change in frequency is approximately:

\frac{\Delta f}{f_0} \approx \frac{v_s}{v} \text{ (for source motion)}

\frac{\Delta f}{f_0} \approx \frac{v_o}{v} \text{ (for observer motion)}

This gives a fast sanity check for your answer.

Common Mistake

Students assume the Doppler effect for sound is symmetric — that a source moving at speed $v_s$ toward a stationary observer gives the same frequency shift as an observer moving at $v_s$ toward a stationary source. This is wrong. The formulas are different: $f_0 \cdot v/(v-v_s)$ versus $f_0 \cdot (v+v_o)/v$ . They give different results (try $v_s = v_o = v/2$ : you get $2f_0$ vs $1.5f_0$ ). The asymmetry exists because the medium (air) breaks the symmetry. NEET 2023 had an MCQ testing exactly this.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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