91Ó°ÊÓ

• Velocity of the wave 343 m/s
• Distance from one source is 19.5m.
• Distance from the other source is 18.3m.

The expression for the velocity of sound in terms of frequency and wavelength is given by,

$v=f\mathrm{λ}$

Here, v is the velocity ,$\mathrm{λ}$ is wavelength and f is the frequency.

1. Condition for destructive interference,$f_{min,n}==\frac{\left(2n-1\right)v}{2∆\text{L}},n=1,2,3,…$
2. Condition for constructive interference,$f_{max,n}=\frac{nv}{∆L},n=1,2,3,…,n=1,2,3,…$

Here is the length.

The phase difference is,

$f=\frac{∆L}{\mathrm{λ}}2\mathrm{π}$

$∆L$is their path length difference.

Fully constructive interference occurs when and destructive interference occurs when,$f=\left(2n+1\right)\mathrm{Ï€}$

Therefore, the condition for destructive interference,

$\frac{∆L}{v}f=n-\frac{1}{2}$……. (i)

Where n = 1,2,3,....

From equation, $v=f\mathrm{λ}$

$\mathrm{λ}=v/f$

Substitute v/f for$\mathrm{λ}$ into the equation (i)

$$\begin{gathered} \frac{∆L}{v}f=n-\frac{1}{2} \\ ∴f=\frac{\left(n-\frac{1}{2}\right)v}{\mathrm{Δ}L} \end{gathered}$$

Now, from the given information, write the path difference as,

$$\begin{gathered} \mathrm{Δ}L=19.5-18.3 \\ =1.2\text{m} \end{gathered}$$

So we can further simplify the equation as,

$$\begin{gathered} f_n=\frac{\left(n-\frac{1}{2}\right)343}{1.2} \\ {f}_n=\left(n-\frac{1}{2}\right)285.833 \end{gathered}$$

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Now, for constructive interference, the condition is,

$f_{max,n}=\frac{mv}{∆L}$

Here n = 1,2,3.....

Substitute 343 m/s for v and$\left(19.5-18.3\right)mfor∆L$ for into the above equation,

role="math" localid="1661326462148" $$\begin{gathered} f_{max,n}=\frac{n×343}{\left(19.5-18.3\right)} \\ {f}_{max,n}=\left(286\right)n \end{gathered}$$……. (iii)

From the above equation (ii), the lowest frequency that gives destructive interference is at n = 1,

$$\begin{gathered} f_n=\left(n-\frac{1}{2}\right)285.833 \\ =142.\text{9Hz} \\ ≈1\text{43Hz} \end{gathered}$$

Therefore the lowest frequency f_min1 that gives minimum signal at the listener’s location is 143 Hz .

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

$$\begin{gathered} f_{min,2}=\left(2-\frac{1}{2}\right)286\text{Hz} \\ {f}_{min,2}=\left(\frac{3}{2}\right)286\text{Hz} \\ {f}_{min,2}=429\text{Hz} \end{gathered}$$

So,

$$\begin{gathered} f_{min,2}=3×143 \\ {f}_{min,2}=3×f_{min,1} \end{gathered}$$

Therefore the number by which f_min1 must be multiplied to get the second lowest frequency f_min2 that gives minimum signal is 3.

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

$$\begin{gathered} f_{min,3}=\left(3-\frac{1}{2}\right)286\text{Hz} \\ =\left(\frac{5}{2}\right)286\text{Hz} \\ =715Hz \end{gathered}$$

Now write f_min3 in the multiple of

$$\begin{gathered} f_{min,3}=5×143 \\ {f}_{min,3}=5×f_{min,1} \end{gathered}$$

Therefore the third lowest frequency f_min3 that gives minimum signal is 715 Hz .

From equation (iii), the lowest frequency that gives constructive interference only when, so that n = 1,

$f_{max,n}=\left(286\right)n$

Substitute 1 for n into the above equation,

$$\begin{gathered} f_{max,1}=\left(286\right)×1 \\ =286\text{Hz} \end{gathered}$$

The lowest frequency f_max1 that gives the maximum signal at the listener’s location is286 Hz.

From equation (iii), the second lowest frequency that gives constructive interference is at n = 2,

$$\begin{gathered} f_{max,n}=\left(286\right)n \\ {f}_{max,2}=2×286\text{Hz} \\ =572\text{Hz} \\ =2×f_{max,1}\text{Hz} \end{gathered}$$

Therefore the number by whichf_max1 must be multiplied to get the second lowest frequency f_max2 that gives maximum signal is 2.

From equation (iii), the third lowest frequency that gives constructive interference is at,

$$\begin{gathered} n=3 \\ {f}_{max,n}=\left(286\right)n \\ {f}_{max,3}=3×286\text{Hz} \\ =858\text{Hz} \\ =3×f_{max,1} \end{gathered}$$

So, the multiplication factor to the f_max1is 3 to get thef_max3 .

Therefore the number by which f_max1 must be multiplied to get the third lowest frequency f_max3 that gives maximum signal is 3.