Problem Statement
A radar operates at a wavelength $\lambda = 3.0\,\text{cm}$. Find the velocity of an approaching aircraft if the beat frequency between the transmitted signal and the signal reflected from the aircraft equals $\Delta\nu = 1.0\,\text{kHz}$.
Given Information
- $\lambda = 3.0\,\text{cm} = 3.0 \times 10^{-2}\,\text{m}$ — radar wavelength
- $\Delta\nu = 1.0\,\text{kHz} = 1.0 \times 10^3\,\text{Hz}$ — beat frequency between transmitted and reflected signals
- $c = 3.0 \times 10^8\,\text{m/s}$ — speed of light
Physical Concepts & Formulas
A radar emits a signal of frequency $\nu_0 = c/\lambda$ and detects the signal reflected from a moving target. Since the aircraft approaches, the reflected signal is Doppler-shifted upward. The shift occurs twice: once as the radar signal hits the approaching aircraft, and again as the reflected signal returns. The beat frequency is therefore twice the single Doppler shift, giving $\Delta\nu = 2v/\lambda$.
- $\Delta\nu = 2v\nu_0/c = 2v/\lambda$ — radar beat frequency for approaching target ($v \ll c$)
- $\nu_0 = c/\lambda$ — frequency of transmitted radar signal
Step-by-Step Solution
Step 1 — Identify the double Doppler shift: The approaching aircraft first receives the radar signal blue-shifted (moving observer), then re-emits it as a moving source. Each encounter shifts frequency by $v\nu_0/c$, so the total beat at the radar is doubled:
$$\Delta\nu = 2\frac{v\nu_0}{c}$$
Step 2 — Express in terms of wavelength: Substitute $\nu_0 = c/\lambda$:
$$\Delta\nu = \frac{2v}{\lambda}$$
Step 3 — Solve for aircraft speed: Rearrange for $v$:
$$v = \frac{\Delta\nu\,\lambda}{2} = \frac{1.0 \times 10^3 \times 3.0 \times 10^{-2}}{2}$$
Worked Calculation
$$v = \frac{1.0 \times 10^3\,\text{Hz} \times 3.0 \times 10^{-2}\,\text{m}}{2} = \frac{30}{2} = 15\,\text{m/s}$$
Answer
$$\boxed{v = 15\,\text{m/s} \approx 54\,\text{km/h}}$$
The aircraft approaches at $15\,\text{m/s}$, a typical landing-approach speed for small aircraft. The radar detects this from a $1\,\text{kHz}$ audio-range beat frequency.
Physical Interpretation
The formula $v = \Delta\nu\,\lambda/2$ is linear in both beat frequency and wavelength. Police radar guns use millimeter-wave bands ($\lambda \sim 10\,\text{mm}$) to resolve small speed differences because shorter wavelength gives larger $\Delta\nu$ for the same $v$. At $15\,\text{m/s}$, the non-relativistic approximation is excellent: the relativistic correction is of order $v^2/c^2 \sim 2.5 \times 10^{-15}$, utterly negligible. Modern Doppler weather radar and air-traffic control systems use exactly this principle, augmented with pulse timing to extract both range and radial velocity simultaneously.
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