Q.Two tuning forks having frequencies 320 Hz and 340 Hz are sounded together to produce sound waves. The velocity of sound in air is 326.4 m s⁻¹. Find the difference in wavelength of these waves.
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The Intuition: When Two Almost-Identical Notes Clash
Imagine you're tuning a guitar. You pluck the string you want to tune, and simultaneously strike the reference note from a tuning fork. If the two notes are exactly the same pitch, you hear a single, steady tone. But if they are almost the same — say one is 440 Hz and the other is 442 Hz — something strange happens. The sound doesn't stay steady. Instead, it swells in loudness, fades away, swells again, fades again, in a slow, throbbing rhythm. That rhythmic pulsing is what we call beats.
Why does this happen? Because the two waves, travelling together, sometimes add up constructively (making a loud sound) and sometimes cancel each other out (making a near-silence). They are constantly shifting in and out of phase.
The Precise Physics: Superposition of Two Slightly Different Frequencies
Let two waves of equal amplitude A but slightly different angular frequencies ω1 and ω2 arrive at your ear. Their displacements at a point can be written as:
y1=Asin(ω1t)
y2=Asin(ω2t)
By the principle of superposition, the resultant displacement is:
y=y1+y2=A[sin(ω1t)+sin(ω2t)]
Using the sum-to-product identity:
sinP+sinQ=2sin(2P+Q)cos(2P−Q)
we get:
y=2Acos(2ω1−ω2t)sin(2ω1+ω2t)
This is the key result. The resultant wave has two parts:
- A fast oscillation at the average frequency 2ω1+ω2 (which is nearly the same as the original frequencies). This is what your ear hears as the pitch.
- A slowly varying amplitude given by 2Acos(2ω1−ω2t). This envelope modulates the loudness.
fbeat=∣f1−f2∣
The beat frequency is the absolute difference of the two original frequencies. Your ear perceives one loud-soft cycle for every complete cycle of the cosine envelope. Since the cosine goes through a full cycle when its argument changes by 2π, the time period of one beat is:
Tbeat=∣ω1−ω2∣/22π=∣ω1−ω2∣2π
And since f=ω/2π, the beat frequency in hertz is simply:
fbeat=∣f1−f2∣
What You Actually Hear
Your ear does not follow the rapid (ω1+ω2)/2 oscillations individually — that's just the pitch you perceive. What you notice is the envelope: the amplitude rises and falls at the beat frequency. So with 440 Hz and 442 Hz, you hear a note of roughly 441 Hz that grows louder and softer 2 times every second. …
λ = v/n for each; Δλ = v(1/n₁ − 1/n₂). …
λ₁ − λ₂ = 326.4(1/320 − 1/340) = 0.06 m.
The two waves travel at the same speed v, so their wavelengths are λ = v/n. With n₁ = 320 Hz < n₂ = 340 Hz, λ₁ > λ₂, and …
Use λ = v/n for each fork and subtract (the lower frequency gi …
Subtracting frequencies instead of wavelengths — the question asks for the differenc …
- CBSE 2024Set ANN1 markQ.What are beats?
›Reveal solutionSolution
Beats are the periodic waxing and waning of sound loudness produced by the superposition of two nearly-equal-frequency sound waves.
When two sound waves of slightly different frequencies (f₁ and f₂, close to each other) but nearly equal amplitude travel in the same direction and superpose, the resultant amplitude at any point does not stay constant — it periodically rises to a maximum and falls to a minimum. This periodic variation in the intensity (loudness) of the resultant sound, heard as a regular waxing and waning, is called beats.
The number of beats produced per second (the beat frequency) equals the difference between the two individual frequencies:
fbeat = |f₁ − f₂|
…
- CBSE 2024Set ANNUAL1 markMCQQ.Which of the following phenomena in sound waves is responsible for the production of beats? (A) Reflection (B) Refraction (C) Interference (D) Resonance
›Reveal solutionSolution
Beats are a periodic interference effect between two close-frequency waves.
When two sound waves of slightly different frequencies f1 and f2 travel through the same medium simultaneously, they superpose (interfere). At some instants their crests align (constructive interference, loud sound) and at others crest meets trough (destru …
- CBSE 2024Set ANNUAL1 markMCQQ.Which property of sound waves is responsible for the formation of beats?(a) Amplitude(b) Wavelength(c) Frequency(d) Speed
›Reveal solutionSolution
Beats are produced when two waves of nearly equal but slightly different frequencies superpose, causing periodic constructive/destructive interference — so it's the difference in FREQUENCY between the two waves that creates beats.
When two sound waves of frequencies f1 and f2 (close in value) overlap, their superposition produces a resultant wave whose amplitude rises and falls periodically. This is heard as a throbbing/pulsing loudness variation called 'beats,' occurring at a beat frequency = |f1 - f2|.
…
- CBSE 2022Set ANNUAL1 markMCQQ.Beats are heard clearly when the frequencies of the two waves are —(a) Equal(b) Nearly equal(c) Very different(d) None of these
›Reveal solutionSolution
Beats are clearly heard when the two frequencies are nearly equal.
Beats are the periodic rise and fall of loudness produced when two waves of slightly different frequencies superpose; the beat frequency equals the difference of the two frequencies.
…
- CBSE 2022Set ANNUAL1 markMCQQ.Beats are produced due to ?(a) Reflection(b) Refraction(c) Interference(d) Resonance of sound waves
›Reveal solutionSolution
Beats are produced by interference of two sound waves of slightly different frequency.
When two waves of nearly equal frequencies travel in the same direction and superpose, they interfere constructively and destructively alternately, producing a periodic variation in intensity called beats …
- CBSE 2020Set ANNUAL1 markMCQQ.Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On sounding these forks together, the time interval between two successive maximum intensities will be _______. (A) 41 s (B) 21 s (C) 1 s (D) 4 s
›Reveal solutionSolution
Beat frequency = |f1 − f2| = 4 Hz, so the time between successive intensity maxima is 1/4 s.
Beat frequency fbeat=∣f1−f2∣=∣454−450∣=4 Hz. The time interval between two successive maxima (loudness peaks) equals the bea …
- CBSE 2019Set ANNUAL1 markQ.When two sound sources are sounded together, then 2 beats are produced in .20 second. Find the frequency of the beats.
›Reveal solutionSolution
Beat frequency = number of beats ÷ time taken = 2 beats / 0.20 s = 10 Hz.
When two sound sources of slightly different frequency are sounded together, the resultant intensity waxes and wanes periodically — this phenomenon is called beats. The beat frequency equals the difference between the two source frequencies, and is measured as the number of beats produced per second: …
- CBSE 2019Set ANNUAL1 markQ.Write the formula of beat frequency.
›Reveal solutionSolution
The beat frequency is simply the magnitude of the difference between the two individual frequencies: f_beat = |f1 - f2|.
When two sound waves of nearly equal but slightly different frequencies f1 and f2 travel together, their superposition produces a periodic waxing and waning of loudness called beats. The number of such loud-soft cycles heard per second (the beat frequency) …
- CBSE 2017Set ANNUAL1 markQ.Define beats.
›Reveal solutionSolution
Beats are the periodic loudness variation produced when two waves of nearly equal frequency interfere; beat frequency = |f1 - f2|.
When two sound waves of slightly different frequencies f1 and f2 (with f1 close to f2) travel through the same medium and superpose, at some instants they arrive in phase and reinforce each other (loud sound), and shortly after they arrive out of phase and cancel each other (faint/no sound) — because the phase difference between them keeps changing steadily with time. This periodic waxing and waning of the resultant sound intensity is called beats. The number of times the loudness becomes maximum per second is the beat frequency, and it equals the difference of the two individual frequencies: f_beat = |f1 - f2|. Beats are commonly used by musicians to tune instruments to the same pitch — as the two freque …
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