The problems are on the sound waves worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Sound is a longitudinal wave of compressions and rarefactions that needs a medium to travel through. These worked answers correct the 20 problems of the Physics 9-10 worksheet on sound waves, from sound needs a medium to clapping with the echo: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the equation, the substitution and the final value with its unit. Remember to halve the total distance for echoes, and convert prefixes such as kHz, MHz and ms before substituting. Rounding differences in the final digit are acceptable.
Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: IGCSE 3.4
Problem 1 Sound needs a medium
Sound is a wave that needs particles to pass the vibrations along. In space there is a vacuum with almost no particles, so the vibrations cannot travel and no sound is transmitted.Problem 2 Types of frequency
(a) infrasound (below 20 Hz). (b) audible. (c) audible (18 000 Hz is below 20 000 Hz). (d) ultrasound (above 20 kHz).Problem 3 Simple travel time
t = d / v = 660 / 330 = 2.0 sProblem 4 Unit conversions
(a) 15 kHz = 15 000 Hz. (b) 2500 Hz = 2.5 kHz. (c) 3 ms = 3 × 10⁻³ s = 0.003 s.Problem 5 Pitch and loudness
B has the higher pitch because it has more waves in the same time (higher frequency). B is also louder because its waves are taller (larger amplitude).Problem 6 Thunderstorm
d = v t = 330 × 4.5 = 1485 m = 1.5 km (2 s.f.)Problem 7 Clap at a cliff
Total distance = 330 × 1.2 = 396 m. The cliff is half of this: d = 198 m.Problem 8 Depth of a lake
d = v t / 2 = 1500 × 0.12 / 2 = 90 mProblem 9 Wavelength of notes
(a) λ = v / f = 330 / 440 = 0.75 m(b) λ = 330 / 20 000 = 0.0165 m = 1.65 cm
Problem 10 Reading an oscilloscope
1. One wave takes 4 divisions, so T = 4 × 2.0 = 8.0 ms = 0.0080 s.2. f = 1 / T = 1 / 0.0080 = 125 Hz
3. The peak is 1.5 divisions above the middle line.
Problem 11 Rail versus air
Rail: t = 1500 / 5000 = 0.30 s. Air: t = 1500 / 330 = 4.5 s. Gap = 4.5 − 0.30 = 4.2 s.Problem 12 Solid or gas?
Ravi confuses particle motion with wave speed. A sound wave is a vibration passed from particle to particle. In steel the particles are packed tightly and strongly bonded, so each passes the vibration on almost at once. In air the particles are far apart and must travel across gaps first. Sound is faster in steel (5000 m/s) than in air (330 m/s).Problem 13 Spot the error
The student forgot that the sound went to the wall and back. 264 m is the total distance. The wall is at half of this: 132 m.Problem 14 Ultrasound wavelength
f = 2.0 × 10⁶ Hz. λ = 1500 / 2.0 × 10⁶ = 7.5 × 10⁻⁴ m = 0.75 mmProblem 15 Depth in a scan
Total distance = 2 × 0.12 = 0.24 m. t = 0.24 / 1500 = 1.6 × 10⁻⁴ s = 160 μsProblem 16 Two cliffs
1. Near cliff: 330 × 1.0 / 2 = 165 m. Far cliff: 330 × 2.5 / 2 = 412.5 m.2. He stands between them, so the cliffs are 165 + 412.5 = 577.5 m apart (about 580 m).
Problem 17 A hunting bat
1. d = 330 × 0.020 / 2 = 3.3 m2. λ = 330 / 50 000 = 0.0066 m = 6.6 mm
3. Waves are reflected strongly only by objects larger than about their wavelength. The short wavelength of ultrasound lets it reflect from small insects.
Problem 18 Speed from a graph
1. v = 400 / 1.21 = 330.6 m/s (331 m/s)2. Gradient = (400 − 100) / (1.21 − 0.30) = 300 / 0.91 = 329.7 m/s (330 m/s)
3. (340 − 329.7) / 340 × 100 = 3.0 % lower
Problem 19 Cracks in steel
1. d = 5000 × 2.0 × 10⁻⁴ / 2 = 0.50 m2. L = 5000 × 3.2 × 10⁻⁴ / 2 = 0.80 m
3. 0.80 − 0.50 = 0.30 m
