E. Waves and Sound · Waves and SoundBeats: Interference in Time
E3.3 E2.2 E2.1About 60 minutes
Learning goals- I can explain beats as the superposition of two sounds with slightly different frequencies.
- I can calculate beat frequency with f_beat = |f₁ − f₂|.
- I can use beats to find an unknown frequency and to tune an instrument.
Words to know
beats
A regular rise and fall in loudness heard when two sounds of slightly different frequencies are played together.
beat frequency
The number of loud pulses per second: f_beat = |f₁ − f₂|.
in tune (unison)
Two sources sounding exactly the same frequency, so no beats are heard.
Wah-wah-wah
Strike two tuning forks that are almost, but not quite, the same, or listen to two guitar strings that are slightly out of tune. You hear a pulsing wah… wah… wah… The sound swells and fades several times a second. These pulses are called beats, and musicians use them every day to tune their instruments.
Where beats come from
Beats are superposition in time. Picture two sound waves arriving at your ear, one at 440 Hz and one at 444 Hz.
- At some moment their compressions arrive together: they are in phase, interfere constructively, and the sound is loud.
- The 444 Hz wave makes 4 more cycles every second than the 440 Hz wave, so it slowly gets ahead. After ⅛ s it is half a cycle ahead: compressions meet rarefactions, the interference is destructive, and the sound is quiet.
- After ¼ s it is a whole cycle ahead, so the waves are back in phase and the sound is loud again.
The number of loud pulses per second is the beat frequency:
f_beat = |f₁ − f₂|
For 440 Hz and 444 Hz: f_beat = 4 Hz, four beats every second. The pitch you hear is about the average of the two frequencies (442 Hz).
Using beats to tune
To tune a guitar string to a 440 Hz reference tone:
- Play both together and listen for beats.
- Adjust the string's tension. If the beats get slower, you are getting closer. If they get faster, you are going the wrong way.
- When the beats disappear, the two frequencies match: the string is in tune.
Beats only tell you the difference between the frequencies, not which one is higher. To find out, change one source slightly in a known direction and see whether the beat frequency goes up or down.
Beats are easiest to hear when the difference is small. When the difference is more than roughly 10 Hz to 15 Hz, the pulses are too fast to count and you hear a rough sound or two separate notes instead.
Worked exampleExample 1: Beat frequency
Two tuning forks, 512 Hz and 508 Hz, are struck together. Find the beat frequency and the time between loud pulses.
- f_beat = |512 − 508| = 4 Hz.
- Time between beats = 1/f_beat = 1 ÷ 4 = 0.25 s.
4 beats per second, one every 0.25 s.
Worked exampleExample 2: Finding an unknown frequency
A piano key and a 262 Hz tuning fork produce 3 beats per second. When the piano string is tightened slightly, the beats increase to 5 per second. What was the piano's original frequency?
- Beats of 3 Hz mean the piano is either 262 + 3 = 265 Hz or 262 − 3 = 259 Hz.
- Tightening a string raises its frequency.
- If it had been 259 Hz, raising it would bring it closer to 262 Hz and the beats would get slower. Instead, they got faster, so it moved away from 262 Hz.
- So the piano was above the fork: 265 Hz. The tuner should loosen the string.
265 Hz; loosen it to get rid of the beats.
Worked exampleExample 3: Two unknowns
Fork A (400 Hz) makes 6 beats/s with fork X. Fork B (410 Hz) makes 4 beats/s with fork X. Find the frequency of X.
- From A: X = 394 Hz or 406 Hz.
- From B: X = 406 Hz or 414 Hz.
- The only value in both lists is 406 Hz.
Just for funWhy did the two tuning forks never make up their minds?
They kept going back and forth, in phase and out of phase, four times a second. Classic beat-ing around the bush!
Hands-on activityHear the beats
You need: two phones (or a phone and a computer) with a free tone-generator app or website, or two identical glass bottles and some water. Keep the volume low.
- Play 440 Hz on one device and 442 Hz on the other, both at the same moderate volume, about 1 m from you. Count the beats in 10 s. Predict first: f_beat = 2 Hz, so about 20 beats in 10 s.
- Change the second tone to 445 Hz. Count again.
- Now set the second tone to an unknown value chosen by a family member (between 435 and 445 Hz). Use beats and the "is it faster or slower when I raise it?" test to find the hidden frequency.
- Bottle version: put the same amount of water in two identical bottles. Blow across both at once (with a helper). Add a few drops of water to one bottle and listen for beats.
What to look for: 440 + 442 Hz gives a slow pulse (2 per second); 440 + 445 Hz gives 5 per second. With the bottles, a tiny change in water level gives slow beats; a bigger change gives faster beats.
Your turn
Two sounds with slightly different frequencies overlap; they drift in and out of phase, so the interference alternates between constructive (loud) and destructive (quiet).
2Find the beat frequency for 256 Hz and 260 Hz.
3A 330 Hz string and a fork give 2 beats/s. What are the two possible fork frequencies?
4Two notes produce 5 beats per second. How much time passes between loud pulses?
5When tuning, the beats get faster as you tighten the string. Should you keep tightening or loosen it?
Loosen: faster beats mean you are moving away from the reference frequency.
6What pitch do you hear when 440 Hz and 446 Hz sound together?
About the average, 443 Hz, pulsing 6 times per second.
More practice: warm-up
1Write the formula for beat frequency.
2Two identical frequencies are played together. How many beats are heard?
3Beats are an example of what kind of wave behaviour?
Interference (superposition).
4A loud pulse happens when the two waves are in phase or out of phase?
In phase (constructive interference).
5f₁ = 500 Hz, f₂ = 497 Hz. Beat frequency?
More practice: on your own
1A violinist hears 4 beats/s between her A string and a 440 Hz tuner. When she loosens the string slightly, the beats slow to 2 beats/s. What was her string's frequency, and what is it now?
Loosening lowers the frequency and the beats got slower, so it moved closer to 440: it was 444 Hz, now 442 Hz.
2Fork X gives 3 beats/s with a 384 Hz fork and 4 beats/s with a 377 Hz fork. Find X.
From 384 Hz: X = 381 or 387 Hz. From 377 Hz: X = 373 or 381 Hz. The value in both lists is 381 Hz.
3A piano tuner hears 2 beats in 1.0 s between a string and a 523 Hz fork. After a small tightening there are 6 beats in 2.0 s. Find the original frequency.
Beats went from 2 Hz to 3 Hz (faster) after raising the frequency, so the string was above the fork: 525 Hz.
4Why can't beats alone tell you which of two sources is higher?
f_beat = |f₁ − f₂| uses the absolute value: 440 vs 443 and 440 vs 437 both give 3 Hz.
5Two flutes play 523 Hz and 520 Hz. The second flutist warms up her flute and her frequency rises by 2 Hz. What is the new beat frequency?
New frequencies 523 and 522 Hz: f_beat = 1 Hz.
Challenge
1A guitar string has a fundamental f = v/(2L). It is 0.65 m long and in tune at 330 Hz. A second string of the same type and tension is accidentally 0.66 m long. Find its frequency and the beat frequency when both are played together.
Same v: v = 2(0.65)(330) = 429 m/s. f = 429 ÷ (2 × 0.66) = 325 Hz. f_beat = 330 − 325 = 5 Hz.
2Explain why the loudness of beats rises and falls smoothly rather than switching suddenly between loud and silent.
The phase difference between the two waves changes gradually, a little each cycle, so the resultant amplitude changes gradually from A₁ + A₂ (in phase) to |A₁ − A₂| (out of phase) and back.
3Two fire-truck sirens sound at 1000 Hz each. One truck is moving toward you and the other is parked. You hear beats at 30 Hz. Which sound is higher, and can you hear these as beats? (Use what you learned: the moving siren's frequency is changed by its motion.)
The approaching siren sounds higher (1030 Hz) because of its motion (the Doppler effect, next lesson). 30 Hz is too fast to hear as separate beats; you would hear a rough, buzzy sound instead of countable pulses.
Quiz
1. Beats are produced by two sounds that have…
2. Two forks of 300 Hz and 306 Hz produce a beat frequency of…
3. A string and a 440 Hz fork give 2 beats/s. The string's frequency could be…
4. While tuning, the beats become slower. This means…
5. A loud pulse in a beat pattern happens when the waves are…
6. Beats are evidence of which principle?
Big idea: expectation E3.3 asks students to explain and illustrate superposition in standing waves and beat frequencies; E2.2 asks for investigations of interference. Beats are interference in time.
How to help: The two-phone activity is quick and convincing. If anyone in the family plays guitar or violin, ask them to demonstrate tuning by ear.
Common difficulties:
- Forgetting that a beat frequency has two possible answers (above and below).
- Reasoning about tightening/loosening: tightening raises the frequency; faster beats mean moving away from the reference.
- In two-fork problems, list both possible values from each fork and look for the one value that appears in both lists.