Respuesta :
Answer:
The intensity is same at [tex]x_{1}=156.0m [/tex] and at [tex]x_{2}=52.0m [/tex]
Explanation:
Let the 2 sources have the same intensity at a distance 'x' from the origin
Since the energy of the wave front emitted will be distributed over a sphere we have
For source 1, located at origin intensity at a distance of 'x' from origin is given by
[tex]e_{1}=\frac{P_{1}}{4\pi x^{2}}[/tex]
For source 2, located at x = +78.0 m intensity at a distance of 'x' from origin is given by
[tex]e_{2}=\frac{P_{2}}{4\pi (x-78)^{2}}[/tex]
Now it is also given that [tex]P_{1}=4P_{2}[/tex]
Thus equating the intensities we get
[tex]\frac{4E_{2}}{4\pi x^{2}}=\frac{E_{2}}{4\pi (x-78)^{2}}[/tex]
Solving the equation for x we get
[tex]\frac{4}{x^{2}}=\frac{1}{(x-78)^{2}}\\\\(\frac{x}{x-78})^{2}=4\\\\\therefore \frac{x}{x-78}=\pm 2[/tex]
Solving using both the values we get
[tex]\therefore \frac{x_{1}}{x_{1}-78}= 2\\\\2x_{1}-156=x_{1}\\\\\therefore x_{1}=156.0m[/tex]
Similarly the other value is obtained as
[tex]\therefore \frac{x_{2}}{x_{2}-78}= -2\\\\-2x_{2}+156=x_{2}\\\\\therefore -3x_{2}=-156.0m\\\\\Rightarrow x_{2}=52.0m[/tex]
