Q3E - University Physics with Modern Physics

Question

A radio transmitting station operating at a frequency of 120 MHz has two identical antennas that radiate in phase. Antenna B is 9.00 m to the right of antenna A. Consider point P between the antennas and along the line connecting them, a horizontal distance x to the right of antenna A. For what values of x will constructive interference occur at point P?

Step-by-Step Solution

Answer

The values of x where constructive interference occur at point P is$\mathrm{x}=0.75 \mathrm{m}, 2.00 \mathrm{m}, 3.25 \mathrm{m}, 4.50 \mathrm{m}, 5.75 \mathrm{m} ,7.00 \mathrm{m}, \mathrm{and} 8.25 \mathrm{m}.$ .

Step 1: Given Data

Frequency generated- $120\hspace{0.33em}\mathrm{MHz}$ 

Distance between antennas- $9.00 \mathrm{m}$ 

Step 2: (a) Concept of constructive and destructive interference.

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,

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

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,

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

Step 3: (b) Determination of the distance where the constructive interference happens is.

The wavelength is calculated as follows,

$\begin{array}{l}\mathrm{\lambda }=\frac{\mathrm{c}}{\mathrm{f}}\\ =\frac{2.998\times {10}^{8 }\mathrm{m}/\mathrm{s}}{120\times {10}^6 \mathrm{Hz}}\\ =2.50 \mathrm{m}\end{array}$

Refer to the image below,

                                                                                 

The path difference is given as,

  

Putting the value of the calculated wavelength,

 

The range in which x must lie is 0 to 9.00 m. Thus, the values of  x is,

$$\begin{gathered} for, \mathrm{m}=0 \\ \mathrm{x}=4.50 \mathrm{m}. \\ \mathrm{for}, \mathrm{m}=+1 \\ \mathrm{x}=+4.50 \mathrm{m}-1.25 \mathrm{m} \\ =3.25 \mathrm{m} \\ \mathrm{for}, \mathrm{m}=+2 \\ \mathrm{x}=+4.50 \mathrm{m}-2.50 \mathrm{m} \\ =2.00 \mathrm{m} \\ \mathrm{for}, \mathrm{m}=+3 \\ \mathrm{x}=4.50 \mathrm{m}-3.75 \mathrm{m} \\ =0.75 \mathrm{m} \\ \mathrm{for}, \mathrm{m}=-1 \\ \mathrm{x}=4.50 \mathrm{m}-1.25 \mathrm{m} \\ =5.75 \mathrm{m} \\ \mathrm{for}, \mathrm{m}=-2 \\ \mathrm{x}=4.50 \mathrm{m}-2.50 \mathrm{m} \\ =7.00 \mathrm{m} \\ for , m=-3 \\ \mathrm{x}=4.50 \mathrm{m} +3.75 \mathrm{m} \\ =8.25 \mathrm{m} \end{gathered}$$

  

 For constructive interference to occur,  $x$  must have the values x = 0.75 m, 2.00 m, 3.25 m, 4.50 m, 5.75 m, 7.00 m, and 8.25 m.