Question

Two speakers, in phase with each other, both put out sound of frequency 260 Hz. A receiver is 2.50 m from one speaker and distance x from the other, where x > 2.50 m. What is the smallest value of x such that the receiver detects maximum destructive interference

Answer 1

Answer: x=3.16m

Step-by-step explanation:

The condition for destructive interference is given by:

∆r = r1 - r2 = (m + 0.5)lamda

Where

Lamda = speed/frequency

= 343/260 = 1.32m

r1. = x

r2 = 2.50m

Then;

X - 2.5 = (m + 0.5)v/f

X - 2.5 = (m + 0.5)1.32

For m = 0 i.e at maximum destructive interference

x = 3.16m