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soldi70 [24.7K]
3 years ago
11

g Light is a traveling wave! The oscillations are oscillations of electric fields. The electric fields oscillate in the y-direct

ion while the light travels in the x direction, so we can create a wave equation Ey(x,t); Ey is the y-component of the electric field, x is position in meters, and t is time in seconds. Consider light with a wave length of 550 nanometers, a wave speed of 3.00 x 108 m/s , and an amplitude of 10 V/m (volts/meters). Construct the traveling wave equation for this light if at t = 0 seconds, the light is at its maximum positive value and is traveling in the negative x-direction. Your equation, Ey(x,t) , should only have the variables x and t (i.e. calculate the values of the relevant traveling wave quantities to three sig figures and use those in the equation you are constructing, do not use the symbols in your final answer).
Physics
1 answer:
Ahat [919]3 years ago
3 0

Answer:

  E_{y}  (x,t)= 10 cos (1.15 10⁻¹¹ x - 3.427 10¹⁵ t)

Explanation:

The general equation of a traveling wave on the x-axis is

        E_{y}(x,t) = E₀ cos (kx -wt)

the amplitude of the wave is E₀ = 10 V / m

the wave number is

         k = 2π /λ

indicate the value of   λ = 550 nm = 550 10⁻⁹ m

we substitute

         k = 2π / 550 10⁻⁹

         k = 1.15 10⁻¹¹ m⁻¹

angular velocity is related to frequency

          w = 2πf

           

the speed of light is related to the wavelength and frequency

           c = λ f

           f = c /λ

we substitute

           w = 2\pi  \frac{c}\lambda }

           w = 2π  3 10⁸/550 10⁻⁹

           w = 3.427 10¹⁵ rad / s

we substitute in the electric field equation

           E_{y}  (x,t)= 10 cos (1.15 10⁻¹¹ x - 3.427 10¹⁵ t)

Let's test the value of this wave for t = 0 at the point x = 0, the wave is worth

           Ey (0,0) = 10

which corresponds to its maximum positive value, therefore this is the expression of the traveling wave

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\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\

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