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5.1.3Waves

Describing Wave Motion — SQA National 5 Physics Revision Notes & Topical Questions

5 min read 5 parts Lesson 3 of 8

📚 SQA National 5 Physics Hub / Topical Revision
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Curriculum Equation Key Wave Speed & Doppler Effect FormulasFast Exam Solution
v = f × λ | f_obs = f × (v ± v_o) / (v ∓ v_s) | T = 1 / f
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Part 1 of 5Wave Features

5.1.3 Describing Wave Motion

These five quantities describe any regular wave and appear in every wave calculation.

Amplitude
  • The maximum displacement of a particle from its rest position;
    • a larger amplitude means the wave carries more energy.
Wavelength (λ)
  • The distance between two matching points on neighbouring waves, such as crest to crest, measured in metres.
Frequency (f)
  • The number of complete waves passing a point each second, measured in hertz.
Period (T)
  • The time for one complete wave to pass, found using T = 1 ÷ f and measured in seconds.
Wave speed (v)
  • The distance the wave travels each second, measured in metres per second.

Part 2 of 5The Wave Equation

The Wave Equation
Equation
  • v = f × λ, so wave speed equals frequency multiplied by wavelength.
Units
  • Use v in m/s, f in Hz and λ in m; convert kHz, MHz or cm into base units before calculating.
Rearranging
  • f = v ÷ λ and λ = v ÷ f, so the equation can be used to find any one of the three quantities.
Constant speed
  • In a given medium the speed is fixed, so raising the frequency forces the wavelength to become shorter.
Distance–time link
  • Because v = d ÷ t as well, the two equations can be combined in multi-step problems.
Changevfλ
Frequency doubled in same mediumSame×2÷2
Wave enters slower medium (f fixed)DecreasesSameDecreases
Amplitude increased onlySameSameSame

Part 3 of 5Reading a Wave Diagram

1
Locate the middle line, which shows the rest position, and measure from it to a crest to find the amplitude.
2
Measure the distance between two successive crests using the scale on the horizontal axis to find the wavelength.
3
On a graph of displacement against time, read the time for one full cycle to find the period.
4
Calculate frequency from f = 1 ÷ T and then use v = fλ to find the wave speed.
5
Check units and significant figures, and state the answer with the correct unit.

Part 4 of 5Worked Example

Question

  • A sound source produces waves of frequency 50 Hz and wavelength 6.8 m in air.
  • Calculate (a) the speed of the waves and (b) the period.

  1. Use v = fλ with f = 50 Hz and λ = 6.8 m.
  2. v = 50 × 6.8 = 340 m/s.
  3. PeriodT = 1 ÷ f = 1 ÷ 50 = 0.020 s.
    What the symbols mean: T = period, the time for one complete wave (s) · f = frequency (Hz)
  4. Checkin 0.020 s the wave travels 340 × 0.020 = 6.8 m, which is one wavelength.

Answer v = 340 m/s and T = 0.020 s.

Part 5 of 5Exam Checklist & Tips

5.1.3 – Key Points for the Exam

  • Amplitude The maximum distance of a vibrating particle from its rest position.
  • Wavelength The distance between identical points on two consecutive waves.
  • Hertz (Hz)
    • The unit of frequency;
      • one wave per second.
  • Period The time taken for one complete wave.
  • Wave equation v = fλ, linking speed, frequency and wavelength.

Common Mistakes

  • Measuring amplitude from a trough to a crest, which gives twice the true amplitude.
  • Forgetting to convert units such as kHz or cm before substituting into v = fλ.
  • Assuming a higher amplitude means a higher wave speed.
  • Writing frequency in seconds or period in hertz.

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Physics 0625 & 9702 SUVAT Mastery Graph Interpretation

Constant Acceleration Equations (SUVAT) & Slopes

v = u + at  |  s = ut + ½at2  |  v2 = u2 + 2as  |  s = ½(u + v)t

  • Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
  • Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
  • Free Fall in vacuum: a = g ≈ 9.81 m/s2 downwards (air resistance negligible).