OlympIQ
Read before you practise · 7 min

Beat Frequency Is the Difference; Loading and Tension Tell You Which Way

Beats appear when two nearby frequencies are heard together, and the beat frequency is their difference. The puzzle is almost always working out which frequency is higher, and that comes from how loading or tension changed the beat rate.

NSEJS · Wave Motion and Doppler's Effect · 8 linked questions · 100% coverage
Know this first

Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.

Distance · Length of the path travelledDisplacement · Straight-line change in positionResultant force · Net force including direction
Study deeper: NCERT Science · Wave Motion and Doppler's Effect

Beat frequency and amplitude pattern

The beat frequency is |f1 - f2|, and the sound heard has the average frequency (f1 + f2)/2 with loudness that rises and falls at the beat frequency. Two sources of 499 Hz and 501 Hz give a 500 Hz tone whose intensity pulses 2 times per second. For waves y1 = 4 sin 500πt and y2 = 5 sin 508πt, the frequencies are 250 and 254 Hz, so 4 beats per second are heard, and the ratio of maximum to minimum intensity is ((A1 + A2)/(A1 - A2))² = (9/1)² = 81 : 1.

Which frequency is higher?

Loading a fork with wax lowers its frequency, filing it raises it, and increasing the tension of a string raises its frequency. Combine this with whether the beats increase or decrease.

Fork A beats 4 times per second with a 288 Hz fork B, and loading A with wax reduces the beats, which means A was above B:
A = 292 Hz

A 512 Hz fork beats 4 times per second with a piano string; when the tension is raised (raising the string frequency) the beats fall to 2, so the string was below the fork, at 508 Hz. If an unknown source gives 4 beats with 250 Hz, it is 254 or 246 Hz;

if its second harmonic gives 5 beats with 513 Hz, then 2
f = 508 Hz and f = 254 Hz

A trickier case: fork A of 256 Hz gives 4 beats with fork B, so B is 252 or 260 Hz. Waxing B drops it to 250 Hz (6 beats), and when only a little wax is removed the beats fall back to 4, which means B has risen to 252 Hz while still carrying some wax. Since removing only part of the wax gets it to 252 Hz, the unwaxed frequency must be higher than 252 Hz, so B is 260 Hz.

Beats from wavelength or tension differences

Waves of wavelengths 2.00 m and 2.02 m with the same speed v have frequencies v/2.00 and v/2.02, so 2 beats per second requires v (1/2.00 - 1/2.02) = 2, giving v = 404 m/s.

For two identical wires with fundamental frequency 600 Hz, frequency is proportional to sqrt(T), so Δf/
f=(1/2)(ΔT/T);f = (1/2)(\Delta T/T);
6 beats per second need ΔT/
T=2x6/600=0.02T = 2 x 6/600 = 0.02

OlympIQ exam lens

  • ✓Beat frequency = |f1 - f2|; heard pitch is the average frequency
  • ✓Maximum : minimum intensity = ((A1 + A2)/(A1 - A2))²
  • ✓Wax or extra load lowers a fork's frequency; more tension raises a string's frequency
  • ✓Change in the beat rate after a known change tells you which frequency is higher
  • ✓For strings f is proportional to sqrt(T), so ΔT/T = 2 Δf/f
OlympIQ trap

Do not pick the unknown frequency without using the direction of the change. A beat frequency of 4 per second only tells you the unknown is 4 Hz above or 4 Hz below the known one; the effect of loading or tension on the beat rate decides which.

Ready to try the pattern?

100% coverage across the 8 linked questions.

Practise 8 linked questions
Sign in to practise — your XP, streak and analysis update as you solve.