Question

The figure shows a loudspeaker A and point C, where a listener is positioned. |AC| = 3.00 m and the angle θ = 42 °. A second loudspeaker B is located somewhere to the left of A. The speakers vibrate out of phase and are playing a 66.0 Hz tone. The speed of sound is 343 m/s. What is the third closest to speaker A that speaker B can be located, so that the listener hears maximum sound?

Answer 1

given

| AC | = 3 m

θ = 42^o

f = 66 Hz

The speed of sound is 343 m/s = v

from figure

BC - AC = / 2

BC - 3 = 343 / 66 x 2

BC - 3 = 2.598

BC = 5.598

using cosine law

cos60 = ( AC² + AB² - BC² ) / 2 x AC x AB

0.5 = 3² + AB² - 5.598² / 2 x 3 x AB

3 x AB = 9 + AB² - 31.343

AB² - 3 AB - 22.343 = 0

by solving the quadratic equation

AB = 6.4591 m

the third closest to speaker A that speaker B can be located,

so that the listener hears maximum sound is AB = 6.4591 m