Two in-phase loudspeakers emit identical 1000 Hz sound waves along the x-axis. What
distance should one speaker be placed behind the other for the sound to have an amplitude
1.5 times that of each speaker alone?
To find the distance between the two speakers, we can use the principle of superposition for waves. When two sound waves overlap, the resulting amplitude is the sum of the amplitudes of the individual waves.
Let's assume that the amplitude of each speaker alone is A.
When the two sound waves align in phase, the resulting amplitude is 1.5A. This means the amplitude of each individual wave is A.
We know that the amplitude of a sound wave is inversely proportional to the distance from the source. In other words, if we double the distance, the amplitude decreases by a factor of 2.
Since the amplitude of each wave is A, we want the resulting amplitude to be 1.5A. This means the distance between the speakers should be such that the resulting amplitude is 1.5 times the amplitude of each individual wave.
Let's assume the distance between the speakers is D.
When the wave from the first speaker reaches the second speaker, it will have traveled a distance of D. The resulting amplitude at the second speaker will be A.
When the wave from the second speaker reaches the first speaker, it will have traveled a distance of 2D. The resulting amplitude at the first speaker will be 1.5A.
Therefore, the distance between the two speakers should be 2D.
To achieve an amplitude of 1.5A, we need the distance traveled by the sound wave from the first speaker to be twice the distance traveled by the sound wave from the second speaker.
Mathematically, we can write:
2D = D * 2
Simplifying, we get:
2D = 2D
This equation is satisfied for any value of D. Therefore, the distance between the two speakers can be any value.