Question

Two sources of sound, $S_1$ and $S_2$, emitting waves of equal wavelength $20.0cm$, are placed with a separation of $20.0cm$ between them. A detector can be moved on a line parallel to $S_1{S}_2$ and at a distance of $20.0cm$ from it. Initially, the detector is equidistant from the two sources. Assuming that the waves emitted by the sources are in phase, find the minimum distance through which the detector should be shifted to detect a minimum of sound.

Answer 1

Step 1: Given

Wavelength of sound sources: $\lambda =20cm$

Distance between sources: $AC=20cm$

Perpendicular distance of detector from $AC$ is $20cm$

Distance by which detector is shifted (assume): $dcm$

Step 2: Formula Used

Path difference for minima is given by $∆x=\frac{\left(2n+1\right)\lambda }{2}$, where $n$ is number of mode and $\lambda$ is wavelength.

Pythagoras theorem: $hypotenus{e}^2=perpendicula{r}^2+bas{e}^2$

Step 3: Find the minimum distance for hearing

Calculate the minimum hearing distance using the formula. For minimum path difference, $n=0$.

$\begin{array}{rcl}∆x_{min}& =& \frac{\left(2n+1\right)\lambda }{2}\\ & =& \frac{\left(2\times 0+1\right)\lambda }{2}\\ & =& \frac{\lambda }{2}\\ & =& \frac{20}{2}\\ & =& 10cm\end{array}$

Find AB using Pythagoras theorem. Base will be $10-d$ because the detector was in the middle before and then moved by a distance of $d$ towards AB.

$$\begin{gathered} hypotenus{e}^2=perpendicula{r}^2+bas{e}^2 \\ \Rightarrow A{B}^2={20}^2+{\left(10-d\right)}^2 \\ \Rightarrow AB=\sqrt{{20}^2+{\left(10-d\right)}^2} \end{gathered}$$

Find BC using Pythagoras theorem. Base will be $10+d$ because the detector was in the middle before and then moved by a distance of $d$ away from BC.

$$\begin{gathered} hypotenus{e}^2=perpendicula{r}^2+bas{e}^2 \\ \Rightarrow B{C}^2={20}^2+{\left(10+d\right)}^2 \\ \Rightarrow BC=\sqrt{{20}^2+{\left(10+d\right)}^2} \end{gathered}$$

Calculate the path difference by AB from BC. Substitute the values of AB and BC using Pythagoras theorem.

$$\begin{gathered} ∆x=BC-AB \\ \Rightarrow 10=\sqrt{{20}^2+{\left(10+d\right)}^2}-\sqrt{{20}^2+{\left(10-d\right)}^2} \\ \Rightarrow d=12.6cm \end{gathered}$$

Hence, the detector should be shifted through $12.6cm$.