Doppler Effect: Step-by-Step Worked Examples (3)

hard • 3 min read
Tags Doppler Effect

Question

A train moves towards a stationary observer at vs=20v_s = 20 m/s while emitting a whistle of frequency f0=600f_0 = 600 Hz. The speed of sound in air is v=340v = 340 m/s. Find the apparent frequency heard by the observer. After the train passes and moves away at the same speed, find the new apparent frequency.

Solution — Step by Step

When the source moves towards a stationary observer:

f′=f0⋅vv−vsf' = f_0 \cdot \frac{v}{v - v_s}

The minus sign in the denominator indicates the source approaches.

f′=600⋅340340−20=600⋅340320=637.5 Hzf' = 600 \cdot \frac{340}{340 - 20} = 600 \cdot \frac{340}{320} = 637.5 \text{ Hz}

So the observer hears about 637.5 Hz as the train approaches — higher pitch.

For source receding: minus becomes plus.

f′=f0⋅vv+vs=600⋅340360≈566.7 Hzf' = f_0 \cdot \frac{v}{v + v_s} = 600 \cdot \frac{340}{360} \approx 566.7 \text{ Hz}

Approaching: pitch goes up (637 > 600 ✓). Receding: pitch goes down (567 < 600 ✓). The drop in pitch as the train passes is what we hear in real life as the “vroom” descends.

Why This Works

The Doppler shift comes from the source compressing wavefronts in the direction of its motion. When the source is approaching, successive wavefronts are emitted closer together, so the observer encounters more crests per second — higher frequency.

The general formula combining moving source and moving observer:

f′=f0⋅v±vov∓vsf' = f_0 \cdot \frac{v \pm v_o}{v \mp v_s}

Top signs for approach (observer towards source, source towards observer), bottom signs for recession. Memorize the convention by reasoning: approaching always increases f′f'.

Quick check trick: if the answer is higher than f0f_0, you have an approach scenario. If lower, recession. If your sign is wrong, the answer will be obviously off — fix the sign.

Alternative Method

Use the wavelength approach. A source moving towards the observer at vsv_s shortens each wavelength by vsT0=vs/f0v_s T_0 = v_s/f_0. New wavelength λ′=λ0−vs/f0=(v−vs)/f0\lambda' = \lambda_0 - v_s/f_0 = (v - v_s)/f_0. The observer’s f′=v/λ′=vf0/(v−vs)f' = v/\lambda' = vf_0/(v - v_s). Same answer.

Students mix up the formula for “moving observer” vs “moving source.” When the observer moves, the formula has ±vo\pm v_o in the numerator. When the source moves, ∓vs\mp v_s is in the denominator. Mixing them up changes the answer drastically. Identify what’s moving first.

Final answer: f′=637.5f' = 637.5 Hz approaching; f′≈566.7f' \approx 566.7 Hz receding.

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