A speaker produces sound waves with amplitude A. The amplitude is doubled to 2A. What happens to the intensity of the sound?
AIt doubles — intensity is proportional to amplitude
BIt quadruples — intensity is proportional to the square of amplitude
CIt increases by √2 — the relationship is through the square root
DIt stays the same — intensity depends on frequency, not amplitude
Intensity is proportional to the square of amplitude: I ∝ A². Doubling the amplitude gives (2A)² = 4A², so intensity quadruples. This quadratic relationship is the central counterintuitive fact of wave energy — many students expect a linear relationship. It's the reason acoustics uses a logarithmic decibel scale: the physical intensity range of human hearing spans many orders of magnitude.
Question 2 Multiple Choice
A point source of sound is 2 m from a listener. The listener moves to 6 m from the source. By what factor does the intensity change?
AIt decreases by a factor of 3 — intensity is inversely proportional to distance
BIt decreases by a factor of 9 — intensity follows the inverse-square law
CIt decreases by a factor of 6 — the total distance traveled by the wave
DIt decreases by a factor of 4 — distance doubled (approximately)
I ∝ 1/r². The distance tripled (from 2 m to 6 m), so intensity changes by 1/3² = 1/9. The inverse-square law is a geometric result: the same power spreads over a sphere of area 4πr², which grows as r². Tripling the distance means the sphere's area increases by 9×, so each unit area receives 1/9 the power.
Question 3 True / False
For a wave traveling along a one-dimensional string, intensity decreases as 1/r² with distance from the source.
TTrue
FFalse
Answer: False
False. The inverse-square law applies to waves radiating isotropically in three dimensions. A wave on a one-dimensional string does not spread over an expanding surface area — energy stays concentrated along the string, so intensity (ignoring damping) remains constant with distance. The 1/r² law is a geometric consequence of 3D spherical spreading, and it only applies when waves genuinely propagate in open three-dimensional space.
Question 4 True / False
Doubling the distance from a point source reduces the wave amplitude to half its original value.
TTrue
FFalse
Answer: True
True. Since I ∝ A² and I ∝ 1/r², combining gives A ∝ 1/r. Amplitude decreases linearly (not quadratically) with distance for a spherical wave in 3D. If you double the distance, A → A/2. This is why sound becomes dramatically quieter as you move away — both amplitude and intensity fall, but intensity falls faster (as 1/r²) while amplitude falls as 1/r.
Question 5 Short Answer
Why is the inverse-square law for intensity described as a 'purely geometric result'?
Think about your answer, then reveal below.
Model answer: Because it follows from geometry alone, not from any property of the wave or medium. A point source radiates fixed total power P uniformly over a sphere of area 4πr². Intensity I = P/(4πr²) must fall as 1/r² because the same energy is spread over an ever-larger surface as r grows. No energy disappears — it just thins out across a larger area.
The key insight is that the 1/r² law is about the geometry of spheres, not about anything special about waves. Any quantity (light, gravity, electrostatic force) that radiates isotropically from a point source into 3D space will follow an inverse-square law for the same reason. Waves in 2D (surface waves) follow 1/r; waves in 1D (strings) don't diminish at all — the dimensionality of the spreading determines the law.