Question

Two speakers create identical

Answer 1

Answer:

2.185 m

Explanation:

We are given that

Frequency of identical sound wave=$f=240 Hz$

Distance of speaker from person 1=$L_1=1.47 m$

Velocity of sound wave,v=$343 m/s$

We have to find the minimum distance to speaker 2 for there to be destructive interference at that spot.

We know that

$v=f\lambda$

$\lambda=\frac{v}{f}=\frac{343}{240}$

$\lambda=1.43 m$

We know that

$L_2-L_1=(n-\frac{1}{2})\lambda$

For minimum distance

n=1

Substitute the value

$L_2-1.47=(1-\frac{1}{2})\times 1.43$

$L_2-1.47=\frac{1}{2}\times 1.43$

$L_2-1.47= 0.715$

$L_2=0.715+1.47=2.185 m$

Hence, the minimum distance to speaker 2 for to be destructive interference at that spot=2.185 m