Q.The speed of a wave in a certain medium is 900 m/s. If 3000 waves pass over a certain point of the medium in 2 minutes, then compute its wavelength.
Concept understanding — Wave Terminology and Basic Relations
A set of precise terms is needed to compare and distinguish any two wave patterns quantitatively. For a transverse wave, the crest is the highest point of the disturbance above the undisturbed (mean) reference level, and the trough is the lowest point below it; for a longitudinal wave the analogous features are the compression (crowded, high-pressure region) and rarefaction (spread-out, low-pressure region). The wavelength λ (SI unit: metre) is the length of one complete, non-repeating section of the wave pattern -- for a transverse wave, the distance between two consecutive crests (or two consecutive troughs); for a longitudinal wave, the distance between two consecutive compressions (or two consecutive rarefactions). The frequency f (SI unit: hertz, Hz) is the number of complete waves that cross a fixed point per second, while the time period T (SI unit: second) is the time taken for exactly one wave to cross that point; because one wave crossing takes 1/f seconds, frequency and period are reciprocals of each other, T=1/f. The amplitude A of a wave is the maximum displacement of the medium from its mean (reference) position -- essentially the height of a crest or the depth of a trough measured from the undisturbed level -- and it is amplitude alone, not wavelength, frequency or wave speed, that distinguishes two otherwise-identical sinusoidal waves of different "size". The wave (or phase) velocity v is the distance the wave pattern itself advances in one second, and combining the definitions above gives the fundamental wave relation v=fλ (also written v=λ/T): a wave with a higher frequency necessarily has a proportionally shorter wavelength for the same speed, and vice versa, so their product -- the speed -- stays fixed for a given medium. Two further quantities restate the same physics in angular form: the angular frequency ω=2π/T=2πf (unit rad/s) counts radians of phase swept out per second, and the (angular) wave number k=2π/λ (unit rad/m) counts radians of phase packed into each metre of the wave pattern; combining these reproduces the same speed relation as v=ω/k.
f=3000/(2×60)=25 Hz; λ=v/f=900/25=36 m.
λ=36 m
Step 1. The frequency is the number of waves crossing a point per second: f=3000 waves/(2 min×60 s/min)=3000/120=25 Hz.
Step 2. Using v=fλ, the wavelength is λ=v/f=900/25=36 m.
λ=36 m
Convert waves-per-2-minutes to frequency in Hz, then apply λ=v/f.
- Forgetting to convert 2 minutes to 120 seconds before dividing.
- CBSE 2026Set ANNUAL1 markMCQQ.A transverse wave moves from medium A to a medium B. In medium A, the velocity of the transverse wave is 500 ms^-1 and the wavelength is 5 m. The frequency and the wavelength of the wave in medium B, when its velocity is 600 ms^-1, are respectively:(a) 120 Hz and 5 m(b) 100 Hz and 5 m(c) 120 Hz and 6 m(d) 100 Hz and 6 m
›Reveal solutionSolution
Frequency is a property of the source and stays the same across media; only speed and wavelength change, so f = 100 Hz stays fixed and the new wavelength is v/f = 6 m.
When a wave travels from one medium into another, its frequency is determined by the source that produces it and does NOT change at the boundary — only the wave speed and wavelength adjust to the new medium's properties, related by v = f*lambda.
Step 1: Find the frequency in medium A.
v_A = f * lambda_A
500 = f * 5
f = 500/5 = 100 Hz
Step 2: Since frequency is unchanged across the boundary, f in medium B is also 100 Hz.
Step 3: Find the wavelength in medium B using v_B = f * lambda_B.
600 = 100 * lambda_B
lambda_B = 600/100 = 6 m
So the frequency stays 100 Hz and the wavelength becomes 6 m in medium B.
✓Final answerThe correct option is (d) 100 Hz and 6 m — frequency is unchanged (100 Hz) across the boundary, and the new wavelength is v_B/f = 600/100 = 6 m.
- CBSE 2024Set ANNUAL1 markMCQQ.A transverse wave moves from a medium A to a medium B. In medium A, the velocity of the transverse wave is 500 ms^-1 and the wavelength is 5 m. The frequency and the wavelength of the wave in medium B when its velocity is 600 ms^-1, respectively are:(a) 120 Hz and 6 m(b) 120 Hz and 5 m(c) 100 Hz and 6 m(d) 100 Hz and 5 m
›Reveal solutionSolution
Frequency does not change when a wave crosses a boundary between media; only wavelength and speed change, related by v = f λ.
In medium A: vA = 500 m/s, λA = 5 m.
Frequency, f = vA / λA = 500 / 5 = 100 Hz.
Since frequency is determined by the source and stays constant as the wave crosses into medium B:
fB = fA = 100 Hz.
In medium B, vB = 600 m/s, so:
λB = vB / fB = 600 / 100 = 6 m.
✓Final answerThe frequency is 100 Hz and the wavelength in medium B is 6 m — option (c).
- CBSE 2022Set ANNUAL1 markMCQQ.A transverse wave moves from a medium A to a medium B. In medium A the velocity of the transverse wave is 500 ms^-1 and the wavelength is 5 m. The frequency and the wavelength of the wave in medium B, when its velocity is 600 ms^-1, respectively are :(a) 120 Hz and 6 m(b) 120 Hz and 5 m(c) 100 Hz and 6 m(d) 100 Hz and 5 m
›Reveal solutionSolution
When a wave passes from one medium to another, its frequency never changes (it is fixed by the source), but its speed changes because the medium is different, and so its wavelength must also change to satisfy v = fλ.
Step 1 — find the frequency using medium A's data:
v_A = 500 ms^-1, λ_A = 5 m
f = v_A / λ_A = 500 / 5 = 100 Hz
Step 2 — the frequency stays the same (100 Hz) as the wave enters medium B, because frequency is determined by the source, not the medium.
Step 3 — find the new wavelength in medium B using its given speed:
v_B = 600 ms^-1
λ_B = v_B / f = 600 / 100 = 6 m
So in medium B, the wave has frequency 100 Hz (unchanged) and wavelength 6 m (changed, because the speed changed).
✓Final answerThe correct option is (c) 100 Hz and 6 m.
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