A pair of speakers separated by 0.700 m are driven by the same oscillator ata fr
ID: 1727020 • Letter: A
Question
A pair of speakers separated by 0.700 m are driven by the same oscillator ata frequency of 675 Hz. Anobserver, originally positioned at one of the speakers, begins towalk along a line perpendicular to the line joining the speakers.(Assume the speed of sound is 345 m/s.) (a) How far must the observer walk before reaching a relativemaximum in intensity?m
(b) How far will the observer be from the speaker when the firstrelative minimum is detected in the intensity?
m (a) How far must the observer walk before reaching a relativemaximum in intensity?
m
(b) How far will the observer be from the speaker when the firstrelative minimum is detected in the intensity?
m
Explanation / Answer
If L is the hypoteneuse of thetriangle and D is the distance between the speakers andx is the distance from the person to the speaker closer tothem then m = L - x for constructive interference Or m + x = L If we square both sides m22 + x2 + 2mx = L2 Using pythag theorem L2 = D2 + x2 m22 + x2 + 2mx = D2 + x2 m22 + 2mx = D2 m2 *0.51112 + 2 * m * 0.5111 *x = 0.7002 0.26122m2 + 1.0222 m x = 0.49 x = ( 0.49 - 0.26122m2 ) / 1.0222 m m is any integer, but x must be a positive number. So we cansee that 0.49 - 0.26122 m2 > 0 1.75 > m so wemust have m = 1 x = ( 0.49 - 0.26122 * 12 ) / 1.0222 * 1 = = 0.223 meters Part b works the same way, except that m can be 0.5 or 1.5. If you calc both you get m = 0.5 x= 1.444 meters m = 1.5 x = 0.138 meters (youuse this one because it's closer to the speaker) Part b works the same way, except that m can be 0.5 or 1.5. If you calc both you get m = 0.5 x= 1.444 meters m = 1.5 x = 0.138 meters (youuse this one because it's closer to the speaker)Related Questions
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