Two loudspeakers separated by a distance of 1.40 m are connected to the same osc

Question

Two loudspeakers separated by a distance of 1.40 m are connected to the same oscillator so as to produce sound waves of the same frequency and phase. The frequency of the oscillator is 2040 Hz.

i) At what angles away from the straight ahead direction (i.e. the perpendicular
bisector of the line joining the two loudspeakers) will the intensity of the sound be
at a maximum?
ii) The connections to one of the loudspeakers are now reversed so as to produce a
phase shift of 180 degrees in the sound waves from this speaker. what are the new
angular directions of the maxima in the sound intensity?

The velocity of sound is 340 m's

Explanation / Answer

let its be alpha. different in path. d*sin alpha = lamda so that we have. (lamda=v*T=v/f=0.167) so alpha=6.8(degrees). -------- d*sin alpha = lamda/2. so that alpha=3.4 (degrees)

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