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 :

whitneytr12

Answer:

1) [tex]x_{1}=x_{n}-d=0.518-0.204=0.314 m[/tex]

2)  [tex]x_{2}=x_{1}-d=0.314-0.204=0.11 m[/tex]

3)  [tex]x_{2}=x_{1}-d=0.518+0.204=0.722 m[/tex]

4)  [tex]x_{2}=x_{1}-d=0.722+0.204=0.926 m[/tex]

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:

[tex]d=\frac{\lambda}{2}[/tex] (1)

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

[tex]v=f\lambda[/tex] (2)

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

[tex]v=2fd[/tex]

[tex]d=\frac{v}{2f}=\frac{343}{2*840}=0.204m[/tex]

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:

[tex]x_{n}=x-d/2=0.62-(0.204/2)=0.518 m[/tex]  

Therefore the nodes will be:

1) [tex]x_{1}=x_{n}-d=0.518-0.204=0.314 m[/tex]

2)  [tex]x_{2}=x_{1}-d=0.314-0.204=0.11 m[/tex]

3)  [tex]x_{2}=x_{1}-d=0.518+0.204=0.722 m[/tex]

4)  [tex]x_{2}=x_{1}-d=0.722+0.204=0.926 m[/tex]

I hope it helps you!

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