Two identical loudspeakers are driven in phase by a common oscillator at 750 Hz

Question

Two identical loudspeakers are driven in phase by a common oscillator at 750 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.)

Minima Distance from
speaker (m)
1st
2nd
3rd
4th
5th
6th
7th

Explanation / Answer

First, lets find the wavelength of the sound

Using v = f, = v/f = .4573 m

We will also need /2 and /4 calculated later, as you will see, so lets get them now.

/2 = .2287 m        /4 = .1143 m

For the solution. If you stand midway between the two speakers that are driven in phase, then the travel length of the sound at the midway points will be equal. That means right in the center you will have constructive interference.

Then, you start to walk a distance "d" to the left and move closer to that speaker. The sound wave from the left speaker travels a distance "d" less than it took to get to the middle. The speaker on the right has to travel a distance "d" more than it had to before. The difference in this travels distance is therefore "2d"

If the distance "2d" is half a wavelength, that is a condition for destructive interference So

if 2d = /2, we have our first minima. That will be wher d = /4, which is .1143 m from the center.

If the center is .6 m, and you move .1143 m toward either speaker, you will have a minima. (Remember this for later)

Then, if you continue to move another same distance d, you will get constructive interference and another maximum like in the middle. If you move another distance d, you will get the next minima.

In summary, the minimum closest to the center is /4 away on either side. Sussessive minima will be an additional /2 away from those points, and so on. We can now calculate all the points of minima between the two speakers from 0 to 1.2 m in distance

Starting at .6 m (the middle), the first minima away from that wil be

.6 + /4    and .6 - /4

.6 + .1143 = .7143 m        .6 - .1143 = .4857 m

Next, continue adding (to the right) or subtracting (to the left) /2 from the successive points..

.7143 + .2287 = .943 m

.943 + .2287 = 1.172 m    The next will be past the speaker, so we can stop here.

In the other direction,

.4857 - .2287 = .257 m

.257 - .2287 = .0283 m   The next will be past the speaker so we can stop here.

In summary, the order of distances from the smallest distance to the largest distance is

1) .0283 m

2) .257 m

3) .4857 m

4) .7143 m

5) .943 m

6) 1.172 m

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