In Fig. 17-53, a point sourceSof sound waves lies near a reflecting wallAB. A sound detectorDintercepts sound ray ${\mathit{R}}_{\mathbf{1}}$traveling directly fromS. It also intercepts sound ray${\mathit{R}}_{\mathbf{2}}$that reflects from the wall such that the angle of incidence ${\mathit{\theta }}_{\mathbf{l}}$is equal to the angle of reflection ${\mathit{\theta }}_{\mathbf{r}}$. Assume that the reflection of sound by the wall causes a phase shift of 0.500l. If the distances are ${\mathbf{d}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}{\mathbf{d}}_{\mathbf{2}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{d}}_{\mathbf{3}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{m}$,what are the (a) lowest and (b) second lowest frequency at whichR1 andR2 are in phase atD?

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

Reflection shifts the phase of the sound wave by$0.500\lambda .$

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,$∆x=n\lambda$ (i)

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

Path difference between direct and reflected wave is given as:

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

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

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

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

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

Substitute the values in the equation (ii) as:

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

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