Question

Two loudspeakers, 4.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.21 m . Assume the speed of sound is 340 m/s.
Part A

What is the frequency of the sound?

Part B

If the frequency is then increased while you remain 0.35m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Answer 1

Answer:

Explanation:

Given that,

Distance between speaker

L = 4.5m

Minimum intensity at L1 = 0.21m

Speed of sound is

V = 340m/s

A. Frequency of sound f?

The path difference Pd

Distance from the first speaker when you are 0.21m away

d1 = 2.25 + 0.21 = 2.46m

Distance from the second speaker when you move 0.21m closer

d2 = 2.25—0.21 = 2.04m

So, path difference is

Pd = ∆d = d1 — d2

Pd = 2.46—2.04 = 0.42m

Using the destructive interference condition

∆d = (m + ½)λ

m = 0,1,2,3....

When m= 0

∆d = ½λ

0.42 = ½λ

λ = 0.84

Then, using wave equation

v = fλ

Then, f = v / λ

f = 340 / 0.84

f = 404.76Hz

B. Incorrect question

If he is to remain at his initial positions then it is 0.21m from the center.

Then,

Constructive interference is given as

∆d = mλ

Where m = 0,1,2,3

So when m= 1

∆d = λ

And we already got the path difference to be 0.42m

So, ∆d = λ

λ = 0.42

So, applying wave equation

V = fλ

F = v/λ

F = 340/0.42

F = 809.52 Hz

But if we are to use the data given in part B

0.35m from the center..

Following the same principle as part A, the path difference will be 0.35

Therefore, since ∆d = λ

Then, λ = 0.35

So, f, = v/λ

F = 340 /0.35

F = 971.43 Hz