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RUDIKE [14]
3 years ago
7

Demuestre que la función de onda: y(x, t) = A e^−i(kx−ωt+φ) satisface la ecuación de onda lineal. (donde i =√−1).

Physics
1 answer:
Alekssandra [29.7K]3 years ago
5 0

Answer:

La ecuación de onda lineal para una función f(x, t) se escribe como:

\frac{d^2f}{dt^2}  = c^2*\frac{d^2f}{dx^2}

Donde c es una constante que depende de la velocidad de propagación de la onda.

En este caso, tenemos:

f(x) = A*e^{-i*(k*x - w*t + \phi)}

Recordar que la derivada de la exponente es tal que:

k(x,y) = A*e^{c*x + n*y}\\dk/dx = c*A*e^{c*x + n*y}

Entonces las derivadas de f van a ser:

\frac{df}{dt} =  (i*w)*f(x)

\frac{d^2f}{dt^2} = (i*w)^2*f(x) = -(w)^2*f(x)

\frac{df}{dx} = (-i*k)*f(x)

\frac{d^2f}{dx^2} = (-i*k)^2*f(x) = -k^2*f(x)

Entonces podemos reescribir la ecuación de onda como:

\frac{d^2f}{dt^2}  = c^2*\frac{d^2f}{dx^2}

-(w)^2*f(x) = c^2*(-k^2*f(x))

Ahora podemos simplemente definir c^2 = (w/k)^2 y vemos que la ecuación de onda se cumple.

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The Bay of Fundy has the greatest tidal ranges on Earth. What can you infer about the Bay of Fundy?
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The Bay of Fundy has the greatest tidal ranges on Earth. What can you infer about the Bay of Fundy?

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A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it
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Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

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