Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity is minimum when speaker 2 is 10 cm behind speaker 1. The intensity increases as speaker 2 is moved forward and first reaches maximum, with amplitude 2a, when it is 30 cm in front of speaker 1. What is What isThe amplitude of the sound (as a multiple of a) if the speakers are placed side by side?

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are

Thus; for both speakers; the wavelength of the sound is:

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

A = 0.765a

3.2x10^7