AS Daily A Level Physics question
Two identical loudspeakers S1 and S2 are driven from the same signal so they start in phase. They are fixed so that at a microphone P the path difference S2P − S1P is 0.25 m. The speed of sound is 340 m s−1. Initially the frequency is 680 Hz and both speakers produce the same displacement amplitude at P. The frequency is then doubled while the geometry is unchanged, and the drive to S2 is reduced so that its displacement amplitude at P halves; S1 is unchanged. Neglect reflections. Compared with I1, the sound intensity at P when only S1 operates, which statement describes the final intensity at P with both speakers operating?
Answer
The correct answer is B.
Correct: B — 2.25 I1, because the waves arrive in phase and amplitudes add before squaring. A assumes intensity increases in direct proportion to amplitude rather than to its square. B is correct: doubling f makes λ = 340/1360 = 0.25 m so the 0.25 m path difference is one wavelength (in phase), giving net amplitude A + 0.5A = 1.5A and intensity (1.5)² I1 = 2.25 I1. C incorrectly treats the waves as still out of phase after doubling f and subtracts amplitudes (0.5A), then squares to get 0.25 I1. D incorrectly adds individual intensities (1 + 0.5²)I1 = 1.25 I1 as if the sources were incoherent, ignoring phase and interference.