AS Daily A Level Physics question
Two identical loudspeakers L1 and L2 are driven by the same signal generator at one frequency. With only L1 on, a microphone at point P measures a sound-pressure amplitude A; with only L2 on (its gain increased), the amplitude at P is 2A. The leads to L2 are reversed so its cone motion is π out of phase with L1. P is exactly midway between the speakers in an anechoic room, so the path lengths to P are equal. Compared with using L2 alone, what is the sound intensity at P when both speakers are on?
Answer
The correct answer is C.
Correct: C — One quarter of that with L2 alone, because at P the out-of-phase waves give net amplitude (2A − A) and intensity ∝ amplitude². A is wrong because out-of-phase superposition still reduces the amplitude at P; unequal amplitudes prevent perfect cancellation but not reduction. B is wrong because intensities do not simply add when waves have phase differences; amplitudes combine first, and here they partially cancel. C is correct because equal path lengths preserve the π phase difference, so the resultant amplitude is 2A − A = A, giving intensity A² compared with 4A² for L2 alone, i.e. a quarter. D is wrong because zero intensity at the midpoint would require equal amplitudes in perfect antiphase; here the stronger source leaves a residual amplitude A.