Wave speed worksheet answers for GCSE physics (free PDF download)

Problem 1 Classify the waves

(a) Transverse. (b) Longitudinal, since the vibration is parallel to the travel direction. (c) Transverse. (d) Transverse.

Problem 2 Reading a wave graph

1. Amplitude = maximum displacement = 3 cm.
2. The pattern repeats every 12 cm (0 cm to 12 cm), so wavelength = 12 cm.

Problem 3 Frequency and period

(a) f = 1 / T = 1 / 0.050 = 20 Hz
(b) T = 1 / f = 1 / 4.0 = 0.25 s

Problem 4 First wave speed

v = f λ = 50 × 0.30 = 15 m/s

Problem 5 Law of reflection

The angle of incidence equals the angle of reflection. The angle of reflection is 35°, and the angle between the incident and reflected rays is 35° + 35° = 70°.

Problem 6 Waves on a rope

f = v / λ = 3.0 / 1.2 = 2.5 Hz

Problem 7 Wi-Fi wavelength

f = 2.4 GHz = 2.4 × 10⁹ Hz
λ = v / f = 3.0 × 10⁸ / 2.4 × 10⁹ = 0.125 m = 12.5 cm

Problem 8 Radio station

f = 96.3 × 10⁶ Hz; λ = 3.0 × 10⁸ / 96.3 × 10⁶ = 3.115 = 3.1 m

Problem 9 A buoy on Lake Victoria

1. f = 20 / 50 = 0.40 Hz
2. T = 1 / 0.40 = 2.5 s
3. v = f λ = 0.40 × 6.0 = 2.4 m/s

Problem 10 Displacement-time graph

1. Amplitude = 4 cm
2. Four complete waves in 1.0 s: T = 0.25 s and f = 4.0 Hz
3. v = f λ = 4.0 × 0.30 = 1.2 m/s

Problem 11 Into shallow water

1. Deep: v = 5.0 × 3.0 = 15 cm/s. Shallow: v = 5.0 × 2.0 = 10 cm/s.
2. The frequency stays at 5.0 Hz; it is set by the source.

Problem 12 Which gap spreads the waves?

The 1 cm gap. Diffraction is greatest when the gap is smaller than or about the same size as the wavelength. The 20 cm gap is ten times the wavelength, so the waves pass through almost unchanged.

Problem 13 Seismic waves

P arrive after 420 / 6.0 = 70 s. S arrive after 420 / 3.5 = 120 s. Difference = 120 − 70 = 50 s.

Problem 14 Spot the error

The wave speed depends on the rope (and its tension), not on the frequency. Since v = f λ with v fixed, doubling f halves the wavelength. The speed stays the same.

Problem 15 Stadium wave

Speed = distance / time = 400 / 25 = 16 m/s.
λ = v / f = 16 / 0.80 = 20 m

Problem 16 Deep and shallow

1. λ = v / f = 0.24 / 6.0 = 0.040 m = 4.0 cm
2. v = f λ = 6.0 × 0.030 = 0.18 m/s
3. 0.18 / 0.24 = 0.75, which is also 3.0 cm / 4.0 cm

Problem 17 A tsunami

1. f = 220 / 200 000 = 1.1 × 10⁻³ Hz; T = 1 / f = 909 s = 15 minutes (15.2 min).
2. The frequency stays 1.1 × 10⁻³ Hz, so λ = 22 / 1.1 × 10⁻³ = 2.0 × 10⁴ m = 20 km. The energy is squeezed into shorter waves, which grow taller.

Problem 18 Two barriers

1. Angle to the normal = 90° − 40° = 50°.
2. The reflected path leaves at 40° to barrier 1. In the triangle formed with the two barriers (angles 90° and 40°), the path meets barrier 2 at 180° − 90° − 40° = 50° to its surface, so the angle of incidence is 90° − 50° = 40°.
3. The final wave travels back parallel to the incoming wave, in the opposite direction.

Problem 19 Rope data

1. 2.0 × 1.50 = 3.0; 3.0 × 1.00 = 3.0; 4.0 × 0.75 = 3.0; 5.0 × 0.60 = 3.0; 6.0 × 0.50 = 3.0. The speed is 3.0 m/s every time.
2. λ is inversely proportional to f. The rope sets the speed, so v = f λ gives a smaller λ when f increases.
3. λ = 3.0 / 7.5 = 0.40 m

Problem 20 Radio and hills

Diffraction is greatest when the obstacle is about the same size as the wavelength or smaller. The 1500 m waves are much longer than the hill (7.5 times), so they bend round it and fill the shadow. The 0.15 m waves are tiny compared with the hill (more than a thousand times smaller), so there is almost no diffraction and a shadow forms. The long-wavelength signal reaches the town more easily.