Question

Two speakers are driven by a common oscillator at 840 Hz and face each other at a distance of 1.24 m. Locate the points along a line joining the two speakers where relative minima of pressure amplitude would be expected. (Use v = 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.)

Answer 1

1)
`x_(1)=x_(n)-d=0.518-0.204=0.314 m`

2)
`x_(2)=x_(1)-d=0.314-0.204=0.11 m`

3)
`x_(2)=x_(1)-d=0.518+0.204=0.722 m`

4)
`x_(2)=x_(1)-d=0.722+0.204=0.926 m`

Step-by-step explanation:

Let's recall that two antinodes are required for a complete wavelength in standing waves which are produced by the facing speakers.

The equation for the spacing between two nodes of a stationaty wave is:

`d=(\lambda)/(2)` (1)

Now, we know that velocity of a sound wave is:

`v=f\lambda` (2)

We can solve the equation 1 for λ and put it on equation 2

`v=2fd`

`d=(v)/(2f)=(343)/(2*840)=0.204m`

The antinode is formed in the half distance between both speakers, so:

x = 1.24/2 = 0.62 m

The location of the node will be:

`x_(n)=x-d/2=0.62-(0.204/2)=0.518 m`

Therefore the nodes will be:

1)
`x_(1)=x_(n)-d=0.518-0.204=0.314 m`

2)
`x_(2)=x_(1)-d=0.314-0.204=0.11 m`

3)
`x_(2)=x_(1)-d=0.518+0.204=0.722 m`

4)
`x_(2)=x_(1)-d=0.722+0.204=0.926 m`

I hope it helps you!