Question

In figure, two speakers separated by the distance${\mathbf{d}}_1\mathbf{=}\mathbf{\text{2.00m}}$are in phase. Assume the amplitudes of the sound waves from the speakers are approximately the same at the listener’s ear at distance${\mathbf{d}}_2\mathbf{=}\mathbf{\text{3.75m}}$directly in front of one speaker. Consider the full audible range for normal hearing, 20Hz to$\mathbf{20}\mathit{K}\mathit{H}\mathit{z}$.

(a)What is the lowest frequency ${\mathbf{f}}_{\max 1}$
that gives minimum signal (destructive interference) at the listener’s ear? By what number must${\mathbf{f}}_{\max 1}$be multiplied to get

(b) The second lowest frequency${\mathbf{f}}_{\min 2}$that gives minimum signal and

(c) The third lowest frequency${\mathbf{f}}_{\min 3}$ that gives minimum signal ?

(d) What is the lowest frequency ${\mathbf{f}}_{\max 1}$that gives maximum signal (constructive interference) at the listener’s ear ? By what number must${\mathbf{f}}_{\max 1}$be multiplied to get

(e) the second lowest frequency${\mathit{f}}_{\mathbf{m}\mathbf{a}\mathbf{x}\mathbf{2}}$that gives maximum signal and

(f) the third lowest frequency ${\mathbf{f}}_{\min 3}$that gives maximum signal?

Short answer

a. Lowest frequency $f_{min,1}$that gives minimum signal at the listener’s location is$f_{min,1}=343\text{Hz}$

b. The number with which $f_{min,1}$must be multiplied to get the second lowest frequency that gives minimum signal is $3,$and$f_{min,2}=1029\text{Hz}.$

c. The number with which $f_{min,1}$must be multiplied to get the second lowest frequencythat gives minimum signal is 5, and$f_{min,3}=171\text{5Hz.}$

d. The lowest frequency $f_{max,1}$that gives maximum signal at the listener’s location is$f_{max,1}=68\text{6Hz}$

e. The number with which $f_{max,1}$must be multiplied to get the second lowest frequency that gives maximum signal is 2, and$f_{max,2}=1372\text{Hz.}$

f. The number with which $f_{max,1}$must be multiplied to get the third lowest frequency $f_{max,3}$that gives maximum signal is 3, and$f_{max,3}=20\text{58Hz.}$

Step-by-step solution

Step 1: Given

• Velocity of the wave is$\text{343m/s}$.
• Distance from one source is

$\begin{array}{l}{L}_1=d_2\\ =3.7\text{5m}\end{array}$

• Distance between two sources is$d_1=2.\text{00m}$

Determining the concept

Use the conditions for destructive and constructive interference to find the frequencies and the multiplier.

Formulae are as follow:

$v=f\lambda$

Condition for destructive interference,

$f_{min,n}=\frac{(2n-1)v}{2\text{ΔL}}$

Condition for constructive interference,

$$\begin{gathered} f_{max,n}=\frac{nv}{\text{Δ}L} \\ n=1,2,3,\dots \end{gathered}$$

Here, v is velocity, f is frequency, Lis the length and$\lambda$ is the wavelength.

(a) Determining the lowest frequency  fmin,1that gives minimum signal at the listener’s ear

The phase difference is,

$f=\frac{\text{Δ}L}{\lambda }2\pi$

$\text{Δ}L$is their path length difference.

Fully constructive interference occurs when$f=2\pi n,$and destructive interference occurs When, $f=(2n+1)\pi$.

Therefore, the condition for destructive interference is,

$$\begin{gathered} \frac{\text{Δ}L}{\lambda }=n-\frac{1}{2} \\ n=1,2,3,\dots . \end{gathered}$$

Fromtheequation,$v=f\lambda$

$\lambda =v/f$

Putting the value in the above equation,

$$\begin{gathered} \frac{\text{Δ}L}{v}f=n-\frac{1}{2} \\ \therefore f=\frac{(n-\frac{1}{2})v}{\Delta L} \end{gathered}$$

First, find the length of the ear from source2,

$\begin{array}{l}{L}_2=\sqrt{L_1^2+d^2}\\ =\sqrt{{3.75}^2+2^2}\\ =4.2\text{5m}\end{array}$

So, that,

$\begin{array}{l}\text{Δ}L=L_2-L_1\\ =4.25-3.75\\ =0.\text{50m}\end{array}$

For,$f_{min}$,

$f_{min,n}=\frac{(n-\frac{1}{2})343}{0.50}$

$f_{min,n}=(n-\frac{1}{2})686$…… (i)

Now, for constructive interference, the condition is,

$f_{max,n}=\frac{nv}{\text{Δ}L}$

Where, $n=1,2,3,\dots .$

$$\begin{gathered} f_{max,n}=\frac{n\times 343}{0.5} \\ {f}_{max,n}=\left(686\right)n \end{gathered}$$…… (ii)

Fromtheabove equation (i), the lowest frequency that gives destructive interference when$n=1$is,

$\begin{array}{c}{f}_{min,n}=(n-\frac{1}{2})686\\ =(1-\frac{1}{2})686\\ =\left(\frac{1}{2}\right)686\\ =3\text{43Hz}\end{array}$

Hence, lowest frequency$f_{min,1}$ that gives minimum signal at the listener’s location is $f_{min,1}=343\text{Hz}$.

(b) Determining the number with which  fmin,1must be multiplied to get the second lowest frequency fmin2  that gives minimum signal

From equation (i), the second lowest frequency that gives destructive interference is at $n=2$,

$\begin{array}{c}{f}_{min,2}=(2-\frac{1}{2})686\text{Hz}\\ =\left(\frac{3}{2}\right)686\text{Hz}\\ =1029\text{Hz}\\ =3\times 343\end{array}$

$\Rightarrow f_{min,2}=3\times f_{min,1}$

So, the multiplication factor to the $f_{min,1}$is 3 to get $f_{min,2}$.

(c) Determining the number with which fmin,1 must be multiplied to get the second lowest frequency fmin2  that gives minimum signal

From equation (ii), the third lowest frequency that gives destructive interference is at $n=3$,

$\begin{array}{c}{f}_{min,3}=(3-\frac{1}{2})686\text{Hz}\\ =\left(\frac{5}{2}\right)686\text{Hz}\\ =1715\text{Hz}\\ =5\times 343\end{array}$

$\Rightarrow f_{min,3}=5\times f_{min,1}$

So, the multiplication factor to the $f_{min,1}$is 5 to get$f_{min,3}$

(d) Determine the lowest frequency fmin,1 that gives maximum signal at the listener’s ear

From equation (ii), the lowest frequency that gives constructive interference is only when $n=1,$

So, that,
$$\begin{gathered} f_{max,n}=\left(686\right)n \\ {f}_{max,1}=68\text{6Hz} \end{gathered}$$

So, the lowest frequency $f_{max,1}$that gives maximum signal at the listener’s location is$f_{max,1}=68\text{6Hz}$
.

(e) Determining the number with which  fmin,1must be multiplied to get the second lowest frequencyfmin2   that gives maximum signal

From equation (ii), the second lowest frequency that gives constructive interference is at

$$\begin{gathered} n=2, \\ \begin{array}{l}{f}_{max,n}=\left(686\right)n\\ f_{max,2}=2\times 686\text{Hz}\\ =137\text{2Hz}\end{array} \end{gathered}$$

$\Rightarrow f_{max,2}=2\times 68\text{6Hz}$

So, the multiplication factor to the $f_{max,1}$is 2 to get $f_{max,2}$.

(f) Determining the number with which  must be multiplied to get the third lowest frequency  that gives maximum signal

From equation (ii), the third lowest frequency that gives constructive interference is at,$n=3$,

$\begin{array}{l}{f}_{max,n}=\left(686\right)n\\ f_{max,3}=3\times 686\text{Hz}\\ =2058\text{Hz}\end{array}$

$\Rightarrow f_{max,3}=3\times 68\text{6Hz}$

So, the multiplication factor to the $f_{max,1}$is 3 to get $f_{max,3}$.