Doppler effect — all four cases for sound, formula selection

Question

A source of sound with frequency 500 Hz moves toward a stationary observer at 34 m/s. If the speed of sound is 340 m/s, find the apparent frequency. What if the observer moves toward a stationary source at the same speed?

(CBSE Class 11 / JEE Main / NEET pattern)

Solution — Step by Step

f' = f_0 \cdot \frac{v \pm v_o}{v \mp v_s}

flowchart TD
    A["Doppler Effect\nfor Sound"] --> B{"Who is moving?"}
    B -->|"Source moves\ntoward observer"| C["f' = f₀ × v/(v - vₛ)\n(frequency increases)"]
    B -->|"Source moves\naway from observer"| D["f' = f₀ × v/(v + vₛ)\n(frequency decreases)"]
    B -->|"Observer moves\ntoward source"| E["f' = f₀ × (v + v₀)/v\n(frequency increases)"]
    B -->|"Observer moves\naway from source"| F["f' = f₀ × (v - v₀)/v\n(frequency decreases)"]

Sign convention: Numerator — add $v_o$ if observer moves toward source. Denominator — subtract $v_s$ if source moves toward observer. Think: “toward = increase in frequency.”

$v_o = 0$ , $v_s = 34$ m/s (toward observer)

f' = 500 \times \frac{340}{340 - 34} = 500 \times \frac{340}{306} = 500 \times 1.111 = \mathbf{555.6 \text{ Hz}}

$v_o = 34$ m/s (toward source), $v_s = 0$

f' = 500 \times \frac{340 + 34}{340} = 500 \times \frac{374}{340} = 500 \times 1.1 = \mathbf{550 \text{ Hz}}

Even though the relative velocity is the same (34 m/s approach), the apparent frequencies are different: 555.6 Hz vs 550 Hz. This is because the Doppler effect for sound depends on the medium — moving the source compresses/stretches the wavelength, while moving the observer changes the rate of wavefront interception. These are physically different processes.

This asymmetry disappears for light (no medium), where only relative velocity matters.

Why This Works

Sound waves travel through a medium at a fixed speed. When the source moves toward the observer, it chases its own waves, compressing them (shorter wavelength → higher frequency). When the observer moves toward the source, they intercept more wavefronts per second (same wavelength, but higher encounter rate).

The compression of wavelength (source motion) is a more efficient frequency booster than the increased interception rate (observer motion), which is why Case 1 gives a higher apparent frequency than Case 2 for the same approach speed.

Alternative Method — Using the Mnemonic “TOP/BOT”

Numerator (TOP) = $v + v_o$ if observer approaches, $v - v_o$ if observer recedes. Denominator (BOT) = $v - v_s$ if source approaches, $v + v_s$ if source recedes.

Memory trick: Toward → frequency goes Up. If someone moves toward, it increases frequency. Observer toward → add in numerator (makes it bigger). Source toward → subtract in denominator (makes it bigger).

In JEE Main, a common variant: source and observer both moving. Just use the full formula $f' = f_0 \cdot \frac{v + v_o}{v - v_s}$ with correct signs. Both moving toward each other: numerator increases, denominator decreases — maximum frequency shift. Both moving in the same direction: partial cancellation.

Common Mistake

The biggest trap: using the Doppler formula for light ( $f' = f_0\sqrt{\frac{1+v/c}{1-v/c}}$ ) in a sound problem, or vice versa. Sound Doppler depends on the medium and distinguishes source vs observer motion. Light Doppler depends only on relative velocity. In NEET/JEE, sound problems use $f' = f_0 \cdot \frac{v \pm v_o}{v \mp v_s}$ , not the relativistic formula. Always check whether the question is about sound or light.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Mnemonic “TOP/BOT”

Common Mistake

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