A sound wave travelling at a speed of 330 m/s produces 20 crests and 20 troughs in 0.1 second. The wavelength of the sound wave is :

Correct — C, 1.65 m. The count is the whole difficulty, and it resolves cleanly: a single complete wave contains one crest and one trough, so twenty crests accompanied by twenty troughs are twenty complete waves, not forty. Twenty waves passing in 0.1 second gives a frequency of 20 ÷ 0.1 = 200 hertz. The wave relation v = fλ then gives the wavelength directly, λ = 330 ÷ 200 = 1.65 m. The item is fair because the two natural ways of counting agree — count the crests alone and you get twenty waves; count crests and troughs as paired features of the same waves and you still get twenty. It is only the reflex of adding the two numbers to forty that goes wrong, and that reading would give 330 ÷ 400 = 0.825 m, which is deliberately not among the options, so a candidate who makes that mistake has no wrong answer to land on and is pushed back to check. Note also that the speed is quoted, not assumed. Sound travels at about 330 to 344 metres per second in air depending on temperature, so 330 m/s is a normal textbook value and the question hands it to you rather than testing it.

• (a)1.1 m — This is 330 ÷ 300, which needs a frequency of 300 Hz — thirty waves in the tenth of a second. Neither the crest count nor the trough count supports that.
• (b)3.3 m — This is 330 ÷ 100, so it halves the frequency to 100 Hz. It comes from treating the twenty crests and twenty troughs as twenty half-waves, that is ten full waves, which undercounts.
• (d)2.2 m — This corresponds to 150 Hz, a frequency no reading of the crest and trough count produces. It is there as a plausible middle value between the other two wrong options.

Any wave is described by three quantities tied together by one relation: the speed v at which the disturbance travels, the frequency f at which the source vibrates, and the wavelength λ, the distance between two consecutive identical points of the wave. They satisfy v = fλ. For sound in a given medium at a given temperature the speed is essentially fixed, so frequency and wavelength move in opposite directions — a high-pitched note is a short wave, a deep note a long one.

Two habits solve this whole family of questions. The first is to convert whatever the stem counts into a number of complete waves before doing any arithmetic; a crest and its neighbouring trough belong to one wave, and papers set the trap by naming both. The second is to remember which quantity the medium fixes: in sound problems the speed is a property of the air and the frequency a property of the source, so it is the wavelength that adjusts. A wording point worth understanding rather than being troubled by: sound in air is a longitudinal wave, and strictly it has compressions and rarefactions rather than crests and troughs. The two vocabularies are used interchangeably because a graph of pressure or displacement against distance for a sound wave does look like a transverse wave, with a compression sitting where a crest would be and a rarefaction where a trough would be. Nothing in the counting changes.

• The wave relation is v = fλ, so λ = v/f.
• One complete wave has exactly one crest and one trough; 20 crests with 20 troughs are 20 waves.
• Twenty waves in 0.1 second gives a frequency of 200 Hz, and 330 ÷ 200 = 1.65 m.
• Sound in air travels at roughly 330-344 m/s depending on temperature, rising as the air gets warmer.
• Sound is a longitudinal wave, its crests and troughs corresponding to compressions and rarefactions.

• Adding the crests and the troughs to get the number of waves; each complete wave already has one of each.
• Assuming the speed of sound rather than reading it from the stem — different papers quote 330, 340 or 344 m/s.
• Muddling frequency with period; here 0.1 second is the total observation time, not the time for one wave.

As a straight substitution into v = fλ, with the count of crests, troughs, oscillations or vibrations in a stated interval doing the real work.

• practice — not a real PYQ

  A source produces 50 complete waves in 2 seconds in a medium where the wave speed is 340 m/s. The wavelength is:

  • (a)6.8 m
  • (b)13.6 m
  • (c)3.4 m
  • (d)27.2 m

  Answer(b) 13.6 m — the frequency is 50 ÷ 2 = 25 Hz, so λ = 340 ÷ 25 = 13.6 m.

• practice — not a real PYQ

  If the frequency of a sound wave in air is doubled while the temperature stays the same, its wavelength will:

  • (a)double
  • (b)halve
  • (c)stay the same
  • (d)become four times as large

  Answer(b) halve — the speed of sound is set by the medium, so with v fixed in v = fλ, doubling the frequency halves the wavelength.