Q34E - University Physics with Modern Physics

Question

Small speakers A and B are driven in phase at 725 Hz by the same audio oscillator. Both speakers start out 4.50 m from the listener, but speaker A is slowly moved away (Fig. E16.34). (a) At what distance d will the sound from the speakers first produce destructive interference at the listener’s location? (b) If A is moved even farther away than in part (a), at what distance d will the speakers next produce destructive interference at the listener’s

Step-by-Step Solution

Answer

The range is   0.518m,0.303m,0.086m,0.126m

STEP 1 Concept of the destructive interference

The path difference is given as $\mathbf{\delta r}\mathbf{=}\left(n+\frac{1}{2}\right)\mathbf{\lambda }$ where, ${\mathbf{r}}_{\mathbf{1}}$ is the path length of the first speaker and ${\mathbf{r}}_{\mathbf{2}}$ is the path length of the second speaker

STEP 2 Rearrangement of position of the minimums with respect to distance from first speaker

The path difference is given by 

$$\begin{gathered} {\mathrm{r}}_2-{\mathrm{r}}_1=\left(\mathrm{n}+\frac{1}{2}\right)\mathrm{\lambda } \\ 1.25-{\mathrm{r}}_2-{\mathrm{r}}_1=\left(\mathrm{n}+\frac{1}{2}\right)\mathrm{\lambda } \\ {\mathrm{r}}_1=\frac{1}{2}\left(1.25-\left(\mathrm{n}+\frac{1}{2}\right)\mathrm{\lambda }\right) \end{gathered}$$

The wavelength of the sound wave is given as 

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}} \\ =\frac{344}{8\times {10}^2} \\ =0.429\mathrm{m} \end{gathered}$$

STEP 3 Plug in wavelength into the r 1 and find all minimums that give r 1 = [0,1.25 m] because we look for points between the speakers only:

Equation is given by $r_1=\frac{1}{2}\left(1.25-\left(n+\frac{1}{2}\right)\lambda \right)$

For n=0 

$$\begin{gathered} {\mathrm{r}}_1=\frac{1}{2}\left(1.25-\left(0+\frac{1}{2}\right)0.429\right) \\ =0.518\mathrm{m} \end{gathered}$$

 

For n = 1

$$\begin{gathered} {\mathrm{r}}_1=\frac{1}{2}\left(1.25-\left(1+\frac{1}{2}\right)0.429\right) \\ =0.303\mathrm{m} \end{gathered}$$

 

For n = 2

$$\begin{gathered} {\mathrm{r}}_1=\frac{1}{2}\left(1.25-\left(2+\frac{1}{2}\right)0.429\right) \\ =0.089\mathrm{m} \end{gathered}$$

 

For n = 3

$$\begin{gathered} {\mathrm{r}}_1=\frac{1}{2}\left(1.25-\left(3+\frac{1}{2}\right)0.429\right) \\ =0.126\mathrm{m} \end{gathered}$$

Therefore, the range is 0.518m,0.303m,0.086m,0.126m