a) La intensidad y el nivel de intensidad sonora en el origen cuando ambos focos emiten simultáneamente. Primero, calculamos las potencias de cada foco a partir de los niveles de intensidad dados.
β = 10 log 10 ( I I 0 ) ⟹ I = I 0 ⋅ 10 β / 10 \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \implies I = I_0 \cdot 10^{\beta/10} β = 10 log 10 ( I 0 I ) ⟹ I = I 0 ⋅ 1 0 β /10 Para el foco F 1 F_1 F 1 , a r 1 ′ = 2 m r_1' = 2 \text{ m} r 1 ′ = 2 m :
I 1 ′ = I 0 ⋅ 10 β 1 / 10 = 1 ⋅ 10 − 12 ⋅ 10 55 / 10 = 10 − 12 ⋅ 10 5.5 = 10 − 6.5 W ⋅ m − 2 I_1' = I_0 \cdot 10^{\beta_1/10} = 1 \cdot 10^{-12} \cdot 10^{55/10} = 10^{-12} \cdot 10^{5.5} = 10^{-6.5} \text{ W} \cdot \text{m}^{-2} I 1 ′ = I 0 ⋅ 1 0 β 1 /10 = 1 ⋅ 1 0 − 12 ⋅ 1 0 55/10 = 1 0 − 12 ⋅ 1 0 5.5 = 1 0 − 6.5 W ⋅ m − 2 La potencia P 1 P_1 P 1 del foco F 1 F_1 F 1 se obtiene de la intensidad a r 1 ′ r_1' r 1 ′ :
I = P 4 π r 2 ⟹ P = I ⋅ 4 π r 2 I = \frac{P}{4\pi r^2} \implies P = I \cdot 4\pi r^2 I = 4 π r 2 P ⟹ P = I ⋅ 4 π r 2 P 1 = I 1 ′ ⋅ 4 π ( r 1 ′ ) 2 = 10 − 6.5 ⋅ 4 π ( 2 m ) 2 = 16 π ⋅ 10 − 6.5 W P_1 = I_1' \cdot 4\pi (r_1')^2 = 10^{-6.5} \cdot 4\pi (2\text{ m})^2 = 16\pi \cdot 10^{-6.5} \text{ W} P 1 = I 1 ′ ⋅ 4 π ( r 1 ′ ) 2 = 1 0 − 6.5 ⋅ 4 π ( 2 m ) 2 = 16 π ⋅ 1 0 − 6.5 W Para el foco F 2 F_2 F 2 , a r 2 ′ = 2 m r_2' = 2 \text{ m} r 2 ′ = 2 m :
I 2 ′ = I 0 ⋅ 10 β 2 / 10 = 1 ⋅ 10 − 12 ⋅ 10 65 / 10 = 10 − 12 ⋅ 10 6.5 = 10 − 5.5 W ⋅ m − 2 I_2' = I_0 \cdot 10^{\beta_2/10} = 1 \cdot 10^{-12} \cdot 10^{65/10} = 10^{-12} \cdot 10^{6.5} = 10^{-5.5} \text{ W} \cdot \text{m}^{-2} I 2 ′ = I 0 ⋅ 1 0 β 2 /10 = 1 ⋅ 1 0 − 12 ⋅ 1 0 65/10 = 1 0 − 12 ⋅ 1 0 6.5 = 1 0 − 5.5 W ⋅ m − 2 La potencia P 2 P_2 P 2 del foco F 2 F_2 F 2 es:
P 2 = I 2 ′ ⋅ 4 π ( r 2 ′ ) 2 = 10 − 5.5 ⋅ 4 π ( 2 m ) 2 = 16 π ⋅ 10 − 5.5 W P_2 = I_2' \cdot 4\pi (r_2')^2 = 10^{-5.5} \cdot 4\pi (2\text{ m})^2 = 16\pi \cdot 10^{-5.5} \text{ W} P 2 = I 2 ′ ⋅ 4 π ( r 2 ′ ) 2 = 1 0 − 5.5 ⋅ 4 π ( 2 m ) 2 = 16 π ⋅ 1 0 − 5.5 W Ahora, calculamos las distancias de cada foco al origen ( 0 , 0 ) (0,0) ( 0 , 0 ) :
r 1 = ( 0 − 0 ) 2 + ( 0 − 3 ) 2 = 0 2 + ( − 3 ) 2 = 3 m r_1 = \sqrt{(0-0)^2 + (0-3)^2} = \sqrt{0^2 + (-3)^2} = 3 \text{ m} r 1 = ( 0 − 0 ) 2 + ( 0 − 3 ) 2 = 0 2 + ( − 3 ) 2 = 3 m r 2 = ( 0 − 4 ) 2 + ( 0 − 0 ) 2 = ( − 4 ) 2 + 0 2 = 4 m r_2 = \sqrt{(0-4)^2 + (0-0)^2} = \sqrt{(-4)^2 + 0^2} = 4 \text{ m} r 2 = ( 0 − 4 ) 2 + ( 0 − 0 ) 2 = ( − 4 ) 2 + 0 2 = 4 m Calculamos la intensidad en el origen debido a cada foco:
I 1 = P 1 4 π r 1 2 = 16 π ⋅ 10 − 6.5 4 π ( 3 ) 2 = 4 9 ⋅ 10 − 6.5 W ⋅ m − 2 I_1 = \frac{P_1}{4\pi r_1^2} = \frac{16\pi \cdot 10^{-6.5}}{4\pi (3)^2} = \frac{4}{9} \cdot 10^{-6.5} \text{ W} \cdot \text{m}^{-2} I 1 = 4 π r 1 2 P 1 = 4 π ( 3 ) 2 16 π ⋅ 1 0 − 6.5 = 9 4 ⋅ 1 0 − 6.5 W ⋅ m − 2 I 2 = P 2 4 π r 2 2 = 16 π ⋅ 10 − 5.5 4 π ( 4 ) 2 = 16 π ⋅ 10 − 5.5 64 π = 1 4 ⋅ 10 − 5.5 W ⋅ m − 2 I_2 = \frac{P_2}{4\pi r_2^2} = \frac{16\pi \cdot 10^{-5.5}}{4\pi (4)^2} = \frac{16\pi \cdot 10^{-5.5}}{64\pi} = \frac{1}{4} \cdot 10^{-5.5} \text{ W} \cdot \text{m}^{-2} I 2 = 4 π r 2 2 P 2 = 4 π ( 4 ) 2 16 π ⋅ 1 0 − 5.5 = 64 π 16 π ⋅ 1 0 − 5.5 = 4 1 ⋅ 1 0 − 5.5 W ⋅ m − 2 La intensidad total I t o t a l I_{total} I t o t a l en el origen, cuando ambos focos emiten simultáneamente, es la suma de las intensidades (asumiendo fuentes incoherentes):
I t o t a l = I 1 + I 2 = 4 9 ⋅ 10 − 6.5 + 1 4 ⋅ 10 − 5.5 I_{total} = I_1 + I_2 = \frac{4}{9} \cdot 10^{-6.5} + \frac{1}{4} \cdot 10^{-5.5} I t o t a l = I 1 + I 2 = 9 4 ⋅ 1 0 − 6.5 + 4 1 ⋅ 1 0 − 5.5 Para sumar, expresamos ambas potencias de diez con el mismo exponente:
I t o t a l = 4 9 ⋅ ( 10 − 1 ⋅ 10 − 5.5 ) + 1 4 ⋅ 10 − 5.5 = ( 4 90 + 1 4 ) ⋅ 10 − 5.5 I_{total} = \frac{4}{9} \cdot (10^{-1} \cdot 10^{-5.5}) + \frac{1}{4} \cdot 10^{-5.5} = \left( \frac{4}{90} + \frac{1}{4} \right) \cdot 10^{-5.5} I t o t a l = 9 4 ⋅ ( 1 0 − 1 ⋅ 1 0 − 5.5 ) + 4 1 ⋅ 1 0 − 5.5 = ( 90 4 + 4 1 ) ⋅ 1 0 − 5.5 I t o t a l = ( 8 180 + 45 180 ) ⋅ 10 − 5.5 = 53 180 ⋅ 10 − 5.5 W ⋅ m − 2 I_{total} = \left( \frac{8}{180} + \frac{45}{180} \right) \cdot 10^{-5.5} = \frac{53}{180} \cdot 10^{-5.5} \text{ W} \cdot \text{m}^{-2} I t o t a l = ( 180 8 + 180 45 ) ⋅ 1 0 − 5.5 = 180 53 ⋅ 1 0 − 5.5 W ⋅ m − 2 Calculando el valor numérico de la intensidad total:
I t o t a l = 53 180 ⋅ 10 − 5.5 ≈ 0.2944 ⋅ 3.162 ⋅ 10 − 6 ≈ 9.31 ⋅ 10 − 7 W ⋅ m − 2 I_{total} = \frac{53}{180} \cdot 10^{-5.5} \approx 0.2944 \cdot 3.162 \cdot 10^{-6} \approx 9.31 \cdot 10^{-7} \text{ W} \cdot \text{m}^{-2} I t o t a l = 180 53 ⋅ 1 0 − 5.5 ≈ 0.2944 ⋅ 3.162 ⋅ 1 0 − 6 ≈ 9.31 ⋅ 1 0 − 7 W ⋅ m − 2 El nivel de intensidad sonora total β t o t a l \beta_{total} β t o t a l en el origen es:
β t o t a l = 10 log 10 ( I t o t a l I 0 ) = 10 log 10 ( 53 180 ⋅ 10 − 5.5 1 ⋅ 10 − 12 ) \beta_{total} = 10 \log_{10} \left( \frac{I_{total}}{I_0} \right) = 10 \log_{10} \left( \frac{\frac{53}{180} \cdot 10^{-5.5}}{1 \cdot 10^{-12}} \right) β t o t a l = 10 log 10 ( I 0 I t o t a l ) = 10 log 10 ( 1 ⋅ 1 0 − 12 180 53 ⋅ 1 0 − 5.5 ) β t o t a l = 10 log 10 ( 53 180 ⋅ 10 6.5 ) \beta_{total} = 10 \log_{10} \left( \frac{53}{180} \cdot 10^{6.5} \right) β t o t a l = 10 log 10 ( 180 53 ⋅ 1 0 6.5 ) β t o t a l = 10 ( log 10 ( 53 180 ) + 6.5 ) \beta_{total} = 10 \left( \log_{10} \left( \frac{53}{180} \right) + 6.5 \right) β t o t a l = 10 ( log 10 ( 180 53 ) + 6.5 ) β t o t a l ≈ 10 ( log 10 ( 0.2944 ) + 6.5 ) ≈ 10 ( − 0.5309 + 6.5 ) ≈ 10 ( 5.9691 ) ≈ 59.7 dB \beta_{total} \approx 10 (\log_{10}(0.2944) + 6.5) \approx 10 (-0.5309 + 6.5) \approx 10 (5.9691) \approx 59.7 \text{ dB} β t o t a l ≈ 10 ( log 10 ( 0.2944 ) + 6.5 ) ≈ 10 ( − 0.5309 + 6.5 ) ≈ 10 ( 5.9691 ) ≈ 59.7 dB b) La distancia al foco F 1 F_1 F 1 del punto situado sobre el segmento que une ambos focos en el que las intensidades generadas por ambos focos son iguales. Primero, calculamos la distancia total d d d entre los focos F 1 ( 0 , 3 ) F_1(0,3) F 1 ( 0 , 3 ) y F 2 ( 4 , 0 ) F_2(4,0) F 2 ( 4 , 0 ) :
d = ( 4 − 0 ) 2 + ( 0 − 3 ) 2 = 4 2 + ( − 3 ) 2 = 16 + 9 = 25 = 5 m d = \sqrt{(4-0)^2 + (0-3)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16+9} = \sqrt{25} = 5 \text{ m} d = ( 4 − 0 ) 2 + ( 0 − 3 ) 2 = 4 2 + ( − 3 ) 2 = 16 + 9 = 25 = 5 m Sea r 1 r_1 r 1 la distancia desde el punto P P P al foco F 1 F_1 F 1 . Como el punto P P P está sobre el segmento que une ambos focos, la distancia desde P P P al foco F 2 F_2 F 2 será r 2 = d − r 1 = 5 − r 1 r_2 = d - r_1 = 5 - r_1 r 2 = d − r 1 = 5 − r 1 . En el punto P P P , las intensidades generadas por ambos focos son iguales: I 1 ( P ) = I 2 ( P ) I_1(P) = I_2(P) I 1 ( P ) = I 2 ( P ) .
P 1 4 π r 1 2 = P 2 4 π r 2 2 \frac{P_1}{4\pi r_1^2} = \frac{P_2}{4\pi r_2^2} 4 π r 1 2 P 1 = 4 π r 2 2 P 2 P 1 r 1 2 = P 2 ( 5 − r 1 ) 2 \frac{P_1}{r_1^2} = \frac{P_2}{(5-r_1)^2} r 1 2 P 1 = ( 5 − r 1 ) 2 P 2 Sustituimos las expresiones para P 1 P_1 P 1 y P 2 P_2 P 2 :
16 π ⋅ 10 − 6.5 r 1 2 = 16 π ⋅ 10 − 5.5 ( 5 − r 1 ) 2 \frac{16\pi \cdot 10^{-6.5}}{r_1^2} = \frac{16\pi \cdot 10^{-5.5}}{(5-r_1)^2} r 1 2 16 π ⋅ 1 0 − 6.5 = ( 5 − r 1 ) 2 16 π ⋅ 1 0 − 5.5 Simplificamos la expresión:
10 − 6.5 r 1 2 = 10 − 5.5 ( 5 − r 1 ) 2 \frac{10^{-6.5}}{r_1^2} = \frac{10^{-5.5}}{(5-r_1)^2} r 1 2 1 0 − 6.5 = ( 5 − r 1 ) 2 1 0 − 5.5 ( 5 − r 1 ) 2 r 1 2 = 10 − 5.5 10 − 6.5 = 10 ( − 5.5 − ( − 6.5 ) ) = 10 1 = 10 \frac{(5-r_1)^2}{r_1^2} = \frac{10^{-5.5}}{10^{-6.5}} = 10^{(-5.5 - (-6.5))} = 10^{1} = 10 r 1 2 ( 5 − r 1 ) 2 = 1 0 − 6.5 1 0 − 5.5 = 1 0 ( − 5.5 − ( − 6.5 )) = 1 0 1 = 10 ( 5 − r 1 r 1 ) 2 = 10 \left( \frac{5-r_1}{r_1} \right)^2 = 10 ( r 1 5 − r 1 ) 2 = 10 Tomamos la raíz cuadrada. Como r 1 r_1 r 1 es una distancia y el punto está en el segmento, r 1 > 0 r_1 > 0 r 1 > 0 y 5 − r 1 > 0 5-r_1 > 0 5 − r 1 > 0 , así que tomamos la raíz positiva:
5 − r 1 r 1 = 10 \frac{5-r_1}{r_1} = \sqrt{10} r 1 5 − r 1 = 10 5 − r 1 = r 1 10 5 - r_1 = r_1 \sqrt{10} 5 − r 1 = r 1 10 5 = r 1 ( 1 + 10 ) 5 = r_1 (1 + \sqrt{10}) 5 = r 1 ( 1 + 10 ) r 1 = 5 1 + 10 r_1 = \frac{5}{1 + \sqrt{10}} r 1 = 1 + 10 5 Calculando el valor numérico:
r 1 ≈ 5 1 + 3.16227766 = 5 4.16227766 ≈ 1.201 m r_1 \approx \frac{5}{1 + 3.16227766} = \frac{5}{4.16227766} \approx 1.201 \text{ m} r 1 ≈ 1 + 3.16227766 5 = 4.16227766 5 ≈ 1.201 m La distancia al foco F 1 F_1 F 1 es aproximadamente 1.20 m 1.20 \text{ m} 1.20 m .
