Question

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?

Short answer

• a)Lowest frequency at which R₁ and R₂ are in phase at D is 39.3Hz.
• b)Second lowest frequency at which R₁and R₂ are in phase at D is 118Hz

Step-by-step solution

The given data

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

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

Understanding the concept of the Doppler Effect

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)

a) Calculation of lowest frequency for both rays being in phase 

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

b) Calculation of second lowest frequency for rays in phase

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 $f=118Hz$.