Question

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them?

Answer 1

Answer:

The smallest separation distance between the speakers is 0.71 m.

Explanation:

Given that,

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave, f = 240 Hz

Let the speed of sound is 343 m/s in air. The speed of sound is given by the formula as :

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{343\ m/s}{240\ Hz}\\\\\lambda=1.42\ m$

To produce destructive interference at a listener standing in front of them,

$d=\dfrac{\lambda}{2}\\\\d=\dfrac{1.42}{2}\\\\d=0.71\ m$

So, the smallest separation distance between the speakers is 0.71 m. Hence, this is the required solution.

Answer 2

The smallest separation distance will be "0.715 m".

According to the question,

• Frequency, f = 240 Hz
• Speed of sound, v = 343 m/s

The wavelength will be:

→ $\lambda = \frac{v}{f}$

By substituting the values, we get

     $= \frac{343}{240}$

     $= 1.43 \ m$

hence,

The smallest separation distance will be:

= $\frac{\lambda}{2}$

= $\frac{1.43}{2}$

= $0.715 \ m$

Thus the above solution is correct.

Learn more about distance here:

https://brainly.com/question/15409465