Doppler Effect — Bat Echolocation (Worked Example)

Hard

Problem

A bat hunting a moth emits 80 kHz ultrasound while flying toward the (stationary) moth at 8.0 m/s. What frequency reaches the moth? (v = 343 m/s)

Given

f = 80,000 Hz
vs = 8.0 m/s (approaching)
v = 343 m/s

Find

The frequency at the moth f'.

Principle

Same Doppler formula at ultrasonic frequencies: f' = f·v/(v − vs).

Formula

f' = f\frac{v}{v - v_s}

f' = fv/(v − vs)

Solution

Substitute: $f' = 80000 \times 343/(343 - 8)$
Denominator: $335$
Calculate: $f' = 80000 \times 343/335 = 81{,}910\ \mathrm{Hz}$

81.9 kHz

Unit check

Hz × (m/s ÷ m/s) = Hz. Dimensionless speed ratio keeps the result a frequency.

What this means physically

81.9 kHz — shifted up 1.9 kHz. Bats detect such shifts in the echo (double Doppler) to judge prey speed — nature's police radar, 50 million years early.

Common mistake

Applying the shift twice for one leg. Source→moth is one Doppler shift; the echo back is a second, separate shift.