Electromagnetism - Oscilloscope

The figure above represents the trace on the screen of cathode ray oscilloscope. The screen is graduated in centimeters. The spot on the screen moves horizontally with a constant speed of 0.5 centimeter/millisecond and the vertical scale is 2 volts/centimeter. The signal is a superposition of two oscillations. Which of the following are most nearly the observed amplitude and frequency of these two oscillations? 

Oscillation 1 Oscillation 2 A. 5V, 250Hz 2.5V, 1000Hz B.  1.5V, 250Hz 3V, 1500Hz  C. 5V, 6Hz 2V, 2Hz  D. 2.5V, 83Hz 1.25V, 500Hz  E. 6.14V, 98Hz                1.35V, 257Hz

(GR9677 #28)

Solution:

The graph shows one big λ consists of 6 small λs.
λ_big ≈ 6 cm
λ_small ≈ 1 cm 

v = λf

Given: v = 0.5 cm/ms
f_big = v/λ_b = 0.5 cm/(6 cm ms) = 1/ (12 ms) = 10³/(12 s) = 83 Hz (Osc. 1)
f_small = v/λ_s = 0.5 cm/(1 cm ms) = 1/ (2 ms) = 10³/(2 s) = 500 Hz (Osc. 2)

Answer: D

Electromagnetism - Superposition

Questions 54-55 concern a plane electromagnetic wave that is a superposition of two independent orthogonal plane waves and can be written as the real part of 

 E = x̂ E₁ exp [i(kz −ωt)] +  ŷ E₂ exp [i(kz − ωt + π)]
where k, ω, E₁ and E₂ are real

If E₂ = E₁, the tip of the electric field vector will describe a trajectory that, as viewed along the z-axis from positive z and looking toward the origin, is a

A. Line at 45^o to the + x-axis
B. Line at 135^o to the + x-axis
C. Clockwise circle
D. Counterclockwise circle
E. Random path

(GR9677 #54)

Solution:

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂ e^{i(kz − ωt +π)} 
E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂ e^{i(kz − ωt)} · e^iπ 

with 
E₂ = E₁ = E
e^iπ = −1

E = E e^{i(kz − ωt)} x̂ − E e^{i(kz − ωt)} ŷ 
E = a x̂ − a ŷ 

tan θ  = a / (−a) = −1

tan 45  = 1
tan 135 = tan 315 = −1

Answer: B


Note: 
e^iϕ = cos ϕ + isin ϕ
e^iπ = cos π + isin π =  −1 + 0 = −1

Electromagnetism - Polarization

Questions 54-55 concern a plane electromagnetic wave that is a superposition of two independent orthogonal plane waves and can be written as the real part of 

E = x̂ E₁ exp [i(kz −ωt)] +  ŷ E₂ exp [i(kz − ωt + π)]

where k, ω, E₁ and E₂ are real

If the plane wave is split and recombined on a screen after the two portions, which are polarized in the x- and y- directions, have traveled an optical path difference of 2π/k, the observed average intensity will be proportional to

A. E₁² + E₂² 
B. E₁² − E₂² 
C. (E₁+ E₂)² 
D. (E₁− E₂)² 
E. 0

(GR9677 #55)

Solution:

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i(kz − ωt +π )} 

Path difference of 2π/k
E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[k(z + 2π/k) − ωt + π)} 

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[kz − ωt)} · e^i3π 
e^i3π = −1

E = x̂ E₁ e^{i(kz − ωt)}  −  ŷ E₂  e^{i(kz − ωt)} 

Intensity in the x- directions, I_x = |E₁|²  
Intensity in the x- directions, I_y = |−E₂|² 

I_total = I_x + I_y = E₁² + E₂²

Answer: A