Two identical loudspeakers are driven in phase by a common oscillator at 780 Hz

Question

Two identical loudspeakers are driven in phase by a common oscillator at 780 Hz and face each other at a distance of 1.20 m. Locate the points along the line joining the two speakers where relative minima of sound pressure amplitude would be expected. (Take the speed of sound in air to be 343 m/s. Choose one speaker as the origin and give your answers in order of increasing distance from this speaker. Enter 'none' in all unused answer boxes.)

Explanation / Answer

Wavelength of the wave produce is L = v/f = 343/780 = 0.439 m

the conditions for minima of sound pressure is x = 0 ,L/2,2L/2,3L/2,4L/2,5L/2............
then

Minima    Distance from
speaker (m)
1st    x = 0

2nd    x = 0.439/2 = 0.2195 m

3rd    x = 0.439 m

4th    x = 0.6585 m

5th    x = 0.878 m

6th    x = 1.0975 m

7th   none

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