Problem 172 - Standing Wave

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standard highschool problem

Lectures by Walter Lewin. They will make you ♥ Physics.

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An electromagnetic plane wave is a traveling wave that can be mathematically described this way: E = Emax * cos(k*x - ω*t + φ). This description applies for a wave travelling in the +x direction and it has a wave number k = 2π/λ, where λ is the wavelength, and an angular frequency ω.

In order to produce a standing wave (a wave that oscillates in time but the troughs and valleys do not change the position) we need to superimpose 2 traveling waves that are moving in opposite directions. These traveling waves need to have the same amplitude and the same frequency and the right phase angle. As the problem is stated, the waves that are superimposing are oscillating in the y-direction and are travelling in the z-direction, so the plane in which they are manifesting is the (y, z) plane.

If we have the following set-up: one electromagnetic plane wave travelling upwards (in the +z direction) and oscillating in the y-direction described by Eup = Emax * cos(k*y - ω*t + φ) and one electromagnetic plane wave travelling downwards (in the -z direction) also oscillating in the y-direction described by Edown = Emax * cos(k*y + ω*t - φ). If we superimpose the two traveling waves we get E = Eup + Edown. If we work the equations a bit we can have the following formulations: Eup = Emax * cos(k*y - (ω*t - φ)) and Edown = Emax * cos(k*y + (ω*t - φ)). If we substitute k*y by α and ω*t - φ by β we get Eup = Emax * cos(α - β) and Edown = Emax * cos(α + β).

This way we get E = Eup + Edown = Emax * cos(α - β) + Emax * cos(α + β), which evaluates as E = 2*Emax * cos(α)*cos(β) or, if we substitute back, E = 2*Emax * cos(k*y)*cos(ω*t - φ).

If φ = π/2 we get E = 2*Emax * cos(k*y)*sin(ω*t). This way we can easily identify Emax as 3/2, k as π/2 and ω as 10^8*π.

So, we can conclude that:

(a) k = 2π/λ = π/2, thus the wavelength is 4m. The absolute refractive index of the medium is n = c/v, where c is the light speed in vacuum which is approx. 3*10^8 m/s and v is the speed of the travelling waves which evaluates to v = λ*f (where f is the frequency). So we have the travelling wave speed v = λ*ω/(2π) = 4 * 10^8 * π / (2π) = 2 * 10^8 m/s. This way we can evaluate the absolute refractive index of the medium in which the travelling waves are manifesting as n = 3/2.

(b) The maximum value of the electric field is achieved when sin(ω*t) = 1, thus E = 3 * cos(π/2 * y). At the position (x, y, z) = (3, 1/2, 2) we get E = 3 * cos(π/4) = 3/√2 ≈ 2.12 V/m.

mdreugen

@lectures by walter lewin for the love of physics is still the best physics book I have read. I have recreated many of your experiments. I am a Mechanical Engineer with a PE stamp including a Master Licensed Plumber in New York. Thank you for all you have done for the community and keep it going you are an amazing person and I admire the passion you have to teach others. I hope this younger generation can realize how many ways there is to learn, and hopefully than can absorb and apply it.

abinantifabz

(i) Wavelength of the standing wave is 4 meters, (π/2 = 2π/wavelength)

(ii) refractive index of the given medium is 3/2
(Refractive index = velocity of the wave in vacuum / velocity of wave in medium

We know that velocity of electromagnetic wave in vacuum is nearly 3 × 10^8

And velocity of electromagnetic wave in medium is 10^8 π/π/2= 2 × 10^8)

(iii) maximum value of electric field is 3/√2
(As the wave doesn't depend on values of x and z, we can substitute y = 0.5)

imchillbro

Hello My name is crystal choi. Now, I’m currently a physics teacher in korea. ln 10 years, I will go to America to study more about phyics and education. Also, I will teach physics to high school students in a fun way like a professor. I’m so happy to live in the same era.

Yajohnjam

LOTS OF LOVE AND RESPECT TO YOU SIR 🙏❤🙏❤🙏❤🙏❤🙏❤

Visibl

Sir I have been struggling with concept of waves and now after your lecture it feels like it was very thank you so much sir for this🥰

roshaniraundale

Sir you got a new Indian subscriber 🇮🇳🇮🇳 जय हिंद

prakhartrip

Solutions:
Wavelength λ = 4 meters
Refractive index n = 1.5
Max E-field at given point, E_pmax = 2.12 V/m

Method:
Given: E(x, y, z, t) = 3*zhat* cos(pi/2 * y) * sin(1e8 pi t)

Generalizing the constants, we get:
E(x, y, z, t) = Emax*cos(k*y)*sin(ω*t)

k tells us the wavelength (k = 2*pi/λ), and ω tells us frequency (ω=2*pi*f).
Wave speed v = λ*f. Solve for λ & f, and put them together to get v = ω/k.

Refractive index n = c/v:
n = c*k/w

Plug in k = π/2 rad/meter, ω = 1e8 π rad/s, and c = 2.998e8 m/s:
λ = 4 meters
n = 1.5

Call the given location, point p. To find the maximum value of E at point p, set the sine term equal to 1 for its maximum value, and plug in the given y-position. This is independent of x and z.
E_pmax = Emax*cos(k*y)
E_pmax = 3 V/m * cos(π/2 rad/m * 0.5 m) = 3/sqrt(2) V/m
E_pmax = 2.12 V/m

carultch

Superb sir.👌
Can you suggest how to learn Physics from scratch to expert for my nephew who is showing interest in physics.Also please recommend some books sir.
Thank you sir..❤

sherlockhomes

We know,
Refractive Index, n = sqrt[u(r) e(r)]
For water, u(r) = 0.99
e(r) = 80
Therefore, n = 9 (approx)
But we know the refractive index of water is 1.3 (approx).
Why???
{The data for u(r) is taken from Wikipedia}
Why is it?
Please Explain.

edutalkindia

Professor do you do any meets and greets or anything like that? I would love to meet you one day

Phymacss

My dear Legend sir #Lewin
I'm just reading ur book *for the love of physics* reading this book makes me think as if I'm doing the experiment while reading, such is the charm of this book

ahangerzgowhar

E(y, t) = 3·cos(πy/2)·sin(10⁸πt) = 3·cos(ky)·sin(ωt)
Where k = 2π/λ is the wave number.

a) From the cosine term:
k = 2π/λ = π/2
So the wavelength λ = 4 m.
From the sine term:
ω = 10⁸ π = 2πf
f = ω/(2π)
Which gives the frequency f = 10⁸ π /(2π) = 50 MHz.

The velocity of the wave:
v = λ·f = 4·50· 10⁶ = 2·10⁸ m/s
Which is 2/3 of the speed of light, c.
Index of refraction = c/v = 3/2 = 1.5

b) E(x, y, z) = E(3, 0.5, 2) = 3·cos(π·0.5/2)·sin(ωt) =
= 3·cos(π/4)·sin(ωt)
cos(π/4) = 1/√2
The maximum value of sin(ωt) = 1
So max. value of the electric field = 3·(1/√2)·1 = 3/√2 V/m.

ulfhaller