91Ó°ÊÓ

The distances,$d_1=10.0m,d_2=20.0mand{d}_3=12.5m.$

Reflection shifts the phase of the sound wave by$0.500\mathrm{λ}.$

We can find the path difference between the direct and reflected wave. Then using the conditions for constructive interference, we can find the lowest and second-lowest frequency at which R1 and R2 are in phase at D.

Formulae:

For constructive interference, the path difference between the waves,$∆x=n\mathrm{λ}$ (i)

The wavelength of the sound wave, $\mathrm{λ}=\frac{v}{f}$ (ii)

Path difference between direct and reflected wave is given as:

$$\begin{gathered} ∆x=\sqrt{{25}^2+{\left(12.5\right)}^2}-\sqrt{{20}^2+{\left(12.5\right)}^2}+0.500\mathrm{λ} \\ =4.36+0.500\mathrm{λ} \end{gathered}$$ ….. (a)

For constructive interference, using equation (i) in the above equation, we get

For n=1 ,
$$\begin{gathered} \mathrm{λ}=4.36+0.500\mathrm{λ} \\ 0.500\mathrm{λ}=4.36 \\ 0.500\frac{v}{f}=4.36 \\ f=0.500×\frac{343}{4.36} \\ f=39.3Hz \end{gathered}$$

Therefore, the lowest frequency at which R₁ and R₂ are in phase at D is. 39.3Hz

Using equation (i) and (a), we get the second-lowest frequency as follows:

For, n = 2

$$\begin{gathered} 4.36+0.500\mathrm{λ}=2\mathrm{λ} \\ 1.500\mathrm{λ}=4.364 \end{gathered}$$

Substitute the values in the equation (ii) as:

$$\begin{gathered} f=1.500×\frac{343}{4.36} \\ f=118Hz \end{gathered}$$

Therefore, the second lowest frequency at which R₁and R₂ are in phase at D is $f=118Hz$.