f9637a0dff-p-696.pdf - 8 Dos ondas armnicas estn descritas por y1 = 4.0 sen(8x 300t y2 = 4.0 sen(8x 300t 2 Cul es la frecuencia de la onda resultante a

f9637a0dff-p-696.pdf - 8 Dos ondas armnicas estn descritas...

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8. Dos ondas armónicas están descritas por: y 1 = 4.0 sen (8x - 300t) y 2 = 4.0 sen (8x - 300t- 2) ¿Cuál es la frecuencia de la onda resultante? a. 300 b. 48 c. 8 d. 0.8 e. 150 9.- En la figura muestra un aparato simple para demostrar la resonancia en tubos y que puede emplearse para medir la rapidez del sonido en el aire usando el método de la resonancia. Un tubo largo, vertical, abierto se conecta por con una tubería flexible a un recipiente cilíndrico con agua y se coloca un diapasón vibratorio de frecuencia conocida cerca de la parte superior. La longitud de la columna de aire, L, se ajusta al mover verticalmente el recipiente. En un experimento con un diapasón de 1080 Hz, y se observan tres resonancias x1 = 6.5 cm, x2 = 22.2 cn y x3 = 37.7 cn por debajo de la parte superior del tubo. Halle el valor de la rapidez del sonido a partir de estos datos. 10. Dos ondas armónicas están descritas por: y 1 = (6cm) sen π (2x + 3t) y y 2 = (6cm) sen π (2x - 3t) La amplitud de la onda resultante para x = 3cm y t = 5 s es: a) 12 cm b) 3 cm c) 6 cm d) 2.5 cm
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  • Fall '08
  • STAFF
  • Onda, Hercio, 6cm, Amplitud, Longitud de onda

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