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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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In an interference pattern, the intensity is Group of answer choices smaller in regions of constructive interference than in reg
Vesnalui [34]

Answer:

The same in both the regions of constructive interference and the regions of destructive interference.

Explanation:

Interference is a phenomenon which occurs when two waves meet while moving along the same medium . The amplitude formed as a result of the interference could be greater, lower, or the same amplitude.

Constructive and destructive interference result from the interaction of waves that are correlated or coherent with each other. This is because arose from the same source or they have the same or nearly the same frequency.

The waves being coherent, arising from the same source and having the same frequency explains why it’s the same in both the regions of constructive interference and the regions of destructive interference.

4 0
4 years ago
An object has a kinetic energy of 275 j and the momentum of magnitude of 25.0 kg m/s. find the speed and mass of the object
lyudmila [28]
<span>Recall formula for Kinetic energy is: KE = 1/2mv^2, where KE = 275J and momentum (which is 25.0 kg m/s) = m*v Therefore substitute for KE and mv in the equation above to get speed => 275 = 0.5 * 25 * v v = 275/12.5 v = 22m/s to get mass m, recall momentum = m*v => 25= 22*m m= 25/22 = 1.3663kg</span>
6 0
4 years ago
In preparation for the final exam, our astronomy study group has reconvened to discuss Mercury's unique orbital properties. They
grigory [225]

Answer:

The only incorrect statement is from student B

Explanation:

The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.

Let's examine student claims using these rotation periods

Student A. The time for 4 turns around the sun is

           t = 4 88

           t = 352 / 58.7 Earth days

In this time I make as many rotations on itself each one with a time to = 58.7 Earth days

           #_rotaciones = t / to

           #_rotations = 352 / 58.7

           #_rotations = 6

therefore this statement is TRUE

student B. the planet rotates 6 times around the Sun

          t = 6 88

          t = 528 s

The number of rotations on itself is

           #_rotaciones = t / to

           #_rotations = 528 / 58.7

           #_rotations = 9

False, turn 9 times

Student C. 8 turns around the sun

           t = 8 88

           t = 704 days

the number of turns on itself is

            #_rotaciones = t / to

            #_rotations = 704 / 58.7

            #_rotations = 12

True

The only incorrect statement is from student B

6 0
3 years ago
A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
4 years ago
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Ludmilka [50]
Straight
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4 0
3 years ago
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