Question

You are listening to a speaker on a tripod in a room with a nice smooth marble floor, as shown in the figure below. The speaker emits sound of wavelength (lambda). The speaker s and your ear e are the same distanceh above the floor. Your ear and the speaker are a distance d = 4 h away from one another.

Not only does your ear sense the sound wave moving along path 1, but your ear also senses the sound wave that is reflected off of the floor along path 2. The angle (theta) at which the sound hits the floor is the same as the angle (theta) at which sound is reflected from the floor.

(a) What is the phase difference (phi) between the two waves when they reach your ear at point e? Express your final answer in terms of the symbols h, (lambda), and numerical constants as needed.

(b) Assuming that h = 1.8 m, what is the highest audible frequency f that will result in completely destructive interference at point e? (Note: humans can hear sounds from 20 Hz to 20 kHz.)

(c) Suppose that when the sound wave reflects off of the marble floor, it picks up a phase shift of (pi)... or, if you prefer, a shift of (lambda)/2. In this case, what is the highest audible frequency f that will result in completely destructive interference at point e?

Answer 1

Path 1 Path 2

a)

In triangle ABD,

AD = Opposite , BD = adjacent = h , AB = hypoteneuse

$tan\theta =\frac{opposite}{adjacent}=\frac{AD}{BD}=\frac{AD}{h}$

$AD=h tan\theta$       EQ-1

Similarly in triangle BCD , using trigonometry

$CD=h tan\theta$    EQ-2

Using equation-1 and equation 2

AD + CD = $h tan\theta + h tan\theta$

AD = 2 h tantheta

4 h = 2 h tantheta

$tan\theta =2$                       Eq-3

In triangle ABD ,

$sec\theta =\frac{hypoteneuse}{adjacent}=\frac{AB}{BD}=\frac{AB}{h}$

$AB = h sec\theta$

Similarly , $BC = h sec\theta$

Length of path-2 = X₂ = AB + BC = $h sec\theta+h sec\theta = 2 h sec\theta$

X₂ = $2h\sqrt{1+tan^2\theta }$                       $since \, sec\theta =\sqrt{1+tan^2\theta }$

tantheta = 2

so      X2 = $2h\sqrt{1+2^2}$ = $2h\sqrt{5}$

length of path-1 = X₁ = 4h

Path difference = $\Delta X = X_{2}-X_{1 }= 2h\sqrt{5}-4h$

$\Delta X = 0.47h$

The formula for phase difference is given as

$\delta = \frac{2\pi \, \Delta X}{\lambda } \, \, where \delta = phase\, difference$

inserting the values we get

$\delta = \frac{(2\times 3.14) (0.47h)}{\lambda } = \frac{2.95h}{\lambda }$