Beats: The Pulsing of Nearly Equal Frequencies
When two sound waves of nearly equal frequencies superpose in time, the ear does not hear two separate tones. Instead, it registers a single pitch whose loudness waxes and wanes periodically. This phenomenon is called beats.
The key results you must master:
- The audible beat frequency equals the full difference ##|f_1 – f_2|##, not half of it.
- The carrier sits at the mean frequency, while the envelope oscillates at half the difference.
- Because loudness follows intensity (amplitude squared), the squared envelope doubles the cycle count — giving the full difference.
- Loading a tuning fork with wax lowers its frequency; this is the classic JEE route to an unknown frequency.
This quiz tests the derivation, the factor-of-two trap, tuning-fork problems, and the limits of the beat phenomenon. Work carefully with signs and detuning conventions.
Q1. Beats are heard when two sound waves of nearly equal frequencies superpose. The audible beat frequency is given by:
Loudness follows intensity, i.e. the square of the envelope amplitude. Squaring a cosine doubles its cycle count, so the audible beat frequency is the full difference ##|f_1 - f_2|##, not half of it.
Q2. Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. The number of loudness maxima heard per second is:
The beat frequency is ##|260 - 256| = 4## Hz, so four maxima are heard each second. The carrier near 258 Hz is not heard as a separate pitch.
Q3. In the superposition ##p(t) = 2A\cos\left(\dfrac{\omega_1-\omega_2}{2}t\right)\cos\left(\dfrac{\omega_1+\omega_2}{2}t\right)##, the first cosine factor represents:
The first cosine varies slowly and acts as the envelope modulating the amplitude; the second cosine is the fast carrier at the mean frequency.
Q4. The carrier frequency of the resultant of two waves of frequencies ##f_1## and ##f_2## is:
The sum-to-product identity places the fast oscillation at the arithmetic mean ##\dfrac{f_1 + f_2}{2}##, while the envelope oscillates at half the difference.
Q5. The beat period (time between successive loudness maxima) for two sources differing by ##\Delta f## is:
Since the audible beat frequency is ##\Delta f##, the time between successive maxima is its reciprocal ##\dfrac{1}{\Delta f}##.
Q6. For beats to be clearly audible, the frequency difference between the two sources should typically be:
Beats are distinct only when the frequencies differ by a few hertz. If the difference is large, the ear resolves two separate tones and no beats are heard.
Q7. A tuning fork of unknown frequency gives 5 beats per second with a standard fork of 512 Hz. The unknown frequency could be:
The unknown frequency satisfies ##|f - 512| = 5##, giving ##f = 507## Hz or ##f = 517## Hz. A loading test is needed to decide which.
Q8. A tuning fork giving 4 beats per second with a standard fork is loaded with wax. The beat frequency decreases. The original frequency of the fork was:
Loading with wax lowers the fork's frequency. If beats decrease, the fork was approaching the standard from above, so its original frequency was higher than the standard.
Q9. When the amplitudes of the two superposing waves are unequal, the beats are:
Unequal amplitudes mean the envelope never fully cancels, so the minima are not true silences. The beat frequency, however, remains ##|f_1 - f_2|##.
Q10. If two sources have exactly the same frequency, the beat frequency is:
With ##f_1 = f_2##, the difference is zero, so there is no waxing and waning — the loudness stays constant.
Q11. Three tuning forks of frequencies 500 Hz, 504 Hz and 508 Hz are sounded together. The beat frequency heard is:
The extreme frequencies differ by ##508 - 500 = 8## Hz, and the intermediate fork's beats coincide with these, so the composite beat frequency is 8 Hz.
Q12. The essential distinction between steady interference and beats is that beats involve:
Steady interference arises from sources of the same frequency with a fixed phase difference; beats require slightly different frequencies so the phase difference drifts with time.
Q13. A sonometer wire and a tuning fork produce 3 beats per second. On slightly increasing the tension of the wire, the beats vanish. The frequency of the fork is:
Increasing tension raises the wire's frequency. Since beats vanish, the wire's frequency rose to match the fork, so the fork was higher than the wire's original frequency.
Q14. The beat phenomenon is used in practice mainly to:
Musicians listen for the disappearance of beats as two notes are brought to the same frequency, which is the standard method of tuning instruments.
Q15. Which statement about the beat phenomenon is a genuine limitation?
Beats are audible only for small frequency differences (a few hertz). For large differences the ear distinguishes two distinct tones, so no beats are perceived.