A2 Daily A Level Physics question
Two identical loudspeakers emit the same steady tone in phase. Point P lies on the perpendicular bisector, equidistant from both, in a space with negligible reflections. Which statement must be true when one loudspeaker is switched off?
Answer
The correct answer is C.
Correct: C — The sound intensity at P becomes one-quarter of its previous value because the resultant amplitude halves. With both in phase and equal distance, displacements add so the amplitude at P with two speakers is 2A; switching one off leaves amplitude A, and intensity scales with amplitude squared, so (A/2A)^2 = 1/4. A misapplies power-based reasoning and ignores that the original amplitude was doubled at P; removing one source reduces intensity to a quarter, not a half. B neglects superposition: distance is unchanged but losing the in-phase contribution halves the amplitude, so intensity must drop. C matches constructive addition of amplitudes and the square relation between intensity and amplitude. D incorrectly compounds effects; halving amplitude alone explains the change, giving one-quarter rather than one-eighth.