Q1E - University Physics with Modern Physics

Question

Two small stereo speakers A and B that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point P starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker B (Fig. E35.1). For what values of x will the sound this person hears be (a) maximally reinforced, (b) cancelled? Limit your solution to the cases where x $\le$ 1.50 m.

                                               

Step-by-Step Solution

Answer

(a)  The values of x will the sound this person hears be maximally reinforced is $150 \mathrm{cm},116 \mathrm{cm}, 82 \mathrm{cm}, 48 \mathrm{cm}, 14 \mathrm{cm}.$

(b) The values of x will the sound this person hears is cancelled is $133 \mathrm{cm}, 99 \mathrm{cm}, 65 \mathrm{cm}, 31 \mathrm{cm}.$

Step 1: (a) Determination of the values of x will the sound this person hears be maximally reinforced.

The path length the person took denoted by x is in the range of 0 ≤ x ≤150 cm.

The sound is reinforced to its maximum when the path difference between the sound waves is an integral multiple of the wavelength. This is also known as the Constructive interference. The condition is,

${\mathrm{r}}_2-{\mathrm{r}}_1=\mathrm{m\lambda },\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{m}=0,\pm 1, \pm 2,...$                                                                 ...(i)

Here, in this problem ${\mathrm{r}}_2$= 150 cm and ${\mathrm{r}}_2$= x cm.

Substitute the values in equation (i),

$$\begin{gathered} 150 \mathrm{cm}-\mathrm{x}=\mathrm{m}(34 \mathrm{cm}) \\ \mathrm{x}=150 \mathrm{cm}-\mathrm{m}(34 \mathrm{cm}) \end{gathered}$$

 

Thus, the values of x are,

$$\begin{gathered} for m=0 \\ \mathrm{x}=150 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} for m=1 \\ x=150\mathrm{cm}-34\mathrm{cm} \\ =116\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{for} \mathrm{m}=2 \\ \mathrm{x}=150\mathrm{cm}-2\left(34\right)\mathrm{cm} \\ =82\mathrm{cm} \end{gathered}$$

$$\begin{gathered} for m=3 \\ x =150\mathrm{cm}-3\left(34\right)\mathrm{cm} \\ =48\mathrm{cm} \end{gathered}$$

$$\begin{gathered} for m=4 \\ x=150\mathrm{cm}-4\left(34\right)\mathrm{cm} \\ =14\mathrm{cm} \end{gathered}$$

Step 2: (b) Determination of the values of x will the sound this person hears is cancelled.

The sound is cancelled when the path difference between the sound waves is a multiple of the odd number of half wavelengths. This is also known as the destructive interference. The condition is,

${\mathrm{r}}_2-{\mathrm{r}}_1=\left(\mathrm{m}+\frac{1}{2}\right)\mathrm{\lambda },\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{m}=0,\pm 1,\pm 2,...$                                                        ...(ii)

Here, in this problem ${\mathrm{r}}_2$= 150 cm and ${\mathrm{r}}_1$= x cm.

Substitute the values in equation (ii),

$$\begin{gathered} 150 \mathrm{cm}-\mathrm{x}=\left(\mathrm{m}+\frac{1}{2}\right)(34 \mathrm{cm}) \\ \mathrm{x}=150 \mathrm{cm}-\left(\mathrm{m}+\frac{1}{2}\right)(34\hspace{0.33em}\mathrm{cm}) \end{gathered}$$ 

$$\begin{gathered} for m=0 \\ \mathrm{x}=133 \mathrm{cm} \\ \mathrm{for} \mathrm{m}=1 \\ \mathrm{x}=150 \mathrm{cm}-\frac{3}{2}\left(34\right) \mathrm{cm} \\ =99 \mathrm{cm} \\ \mathrm{for} \mathrm{m}=2 \\ \mathrm{x}=150 \mathrm{cm}-\frac{5}{2}\left(34\right) \mathrm{cm} \\ =65 \mathrm{cm} \\ \mathrm{for} \mathrm{m}=3 \\ \mathrm{x}=150 \mathrm{cm}-\frac{7}{2}\left(34\right) \mathrm{cm} \\ =31 \mathrm{cm} \end{gathered}$$ 

Thus, The values of x for which  the sound will be cancelled is 133 cm, 99 cm, 65 cm, 31 cm.