91影视

• 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:

$v=f位$

Condition for destructive interference,

$f_{min,n}=\frac{(2n鈭�1)v}{2\text{螖尝}}$

Condition for constructive interference,

$$\begin{gathered} f_{max,n}=\frac{nv}{\text{螖}L} \\ n=1,2,3,鈥� \end{gathered}$$

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

The phase difference is,

$f=\frac{\text{螖}L}{位}2蟺$

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

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

Therefore, the condition for destructive interference is,

$$\begin{gathered} \frac{\text{螖}L}{位}=n鈭�\frac{1}{2} \\ n=1,2,3,鈥�. \end{gathered}$$

Fromtheequation,$v=f位$

$位=v/f$

Putting the value in the above equation,

$$\begin{gathered} \frac{\text{螖}L}{v}f=n鈭�\frac{1}{2} \\ 鈭�f=\frac{(n鈭�\frac{1}{2})v}{螖L} \end{gathered}$$

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,

$\begin{array}{l}\text{螖}L=L_2鈭�L_1\\ =4.25鈭�3.75\\ =0.\text{50m}\end{array}$

For,$f_{min}$,

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

$f_{min,n}=(n鈭�\frac{1}{2})686$鈥︹�� (i)

Now, for constructive interference, the condition is,

$f_{max,n}=\frac{nv}{\text{螖}L}$

Where, $n=1,2,3,鈥�.$

$$\begin{gathered} f_{max,n}=\frac{n脳343}{0.5} \\ {f}_{max,n}=\left(686\right)n \end{gathered}$$鈥︹�� (ii)

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

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

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

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

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

$鈬�f_{min,2}=3脳f_{min,1}$

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 $n=3$,

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

$鈬�f_{min,3}=5脳f_{min,1}$

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

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

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

$鈬�f_{max,2}=2脳68\text{6Hz}$

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,$n=3$,

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

$鈬�f_{max,3}=3脳68\text{6Hz}$

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