A brief note on Beats, Resonance and the Doppler Effect . Read it before a test. Try each question at the end before you open its answer.
Beats Two close frequencies drift in and out of step. Beat frequency: f b = ∣ f 1 − f 2 ∣ f_b = |f_1 - f_2| f b = ∣ f 1 − f 2 ∣ . Beat period: T b = 1 / f b T_b = 1/f_b T b = 1/ f b . Pitch heard: f 1 + f 2 2 \dfrac{f_1 + f_2}{2} 2 f 1 + f 2 . Wax lowers a fork's frequency. Filing raises it. Tightening raises a string's. Beats faster after a change: the two frequencies moved apart. The resonance tube Node at the water, antinode a distance e e e above the open end. l 1 + e = λ / 4 l_1 + e = \lambda/4 l 1 + e = λ /4 , l 2 + e = 3 λ / 4 l_2 + e = 3\lambda/4 l 2 + e = 3 λ /4 : only odd quarter waves. Speed of sound: v = 2 f ( l 2 − l 1 ) v = 2f(l_2 - l_1) v = 2 f ( l 2 − l 1 ) . The end correction cancels. End correction: e = l 2 − 3 l 1 2 ≈ 0.3 d e = \dfrac{l_2 - 3l_1}{2} \approx 0.3d e = 2 l 2 − 3 l 1 ≈ 0.3 d . The Doppler effect f ′ = f v ± v o v ∓ v s f' = f\,\dfrac{v \pm v_o}{v \mp v_s} f ′ = f v ∓ v s v ± v o . Upper signs when that body moves towards the other. Source moving: the wavelength changes. Listener moving: the rate of meeting crests changes. Speeds are measured relative to the air. Echo from a reflector coming at u u u : f v + u v − u f\,\dfrac{v + u}{v - u} f v − u v + u . A worked example A fire engine with a 600 Hz 600\ \text{Hz} 600 Hz siren passes you at 15 m/s 15\ \text{m/s} 15 m/s .
Coming. f ′ = 600 Hz × 340 m/s 325 m/s = 628 Hz f' = 600\ \text{Hz} \times \dfrac{340\ \text{m/s}}{325\ \text{m/s}} = 628\ \text{Hz} f ′ = 600 Hz × 325 m/s 340 m/s = 628 Hz . Going. f ′ = 600 Hz × 340 m/s 355 m/s = 575 Hz f' = 600\ \text{Hz} \times \dfrac{340\ \text{m/s}}{355\ \text{m/s}} = 575\ \text{Hz} f ′ = 600 Hz × 355 m/s 340 m/s = 575 Hz . The drop. About 53 Hz 53\ \text{Hz} 53 Hz , heard the moment it passes. Practice with new numbers
A program to try The program finds the two notes of the fire engine.
Run it. Check both lines against the worked example.
1. Change s = 15.0 to s = 30.0. Does the drop double?
Mistakes to avoid Adding frequencies for beats. Take the difference. Taking l 2 − l 1 l_2 - l_1 l 2 − l 1 as λ \lambda λ . It is λ / 2 \lambda/2 λ /2 . Swapping the Doppler signs. Towards always raises the note. A one-step echo. The reflector is a listener, then a source. Check yourself Forks of 315 Hz 315\ \text{Hz} 315 Hz and 318 Hz 318\ \text{Hz} 318 Hz sound together. What is the beat period? Stuck? Show a hint Show the answerA tube resounds at l 1 l_1 l 1 and l 2 l_2 l 2 . How much is l 2 − l 1 l_2 - l_1 l 2 − l 1 , in wavelengths? Stuck? Show a hint Show the answerA siren comes towards you, then goes away. When does its note drop? Stuck? Show a hint Show the answerWhy does an echo from a car coming towards you sound higher than the sound that was sent? Stuck? Show a hint Show the answer