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?

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

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

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.

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 $f_{min,1}=343\text{Hz}$.

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 $f_{min,2}$.

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$f_{min,3}$

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$f_{max,1}=68\text{6Hz}$
.

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 $f_{max,2}$.

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 $f_{max,3}$.