Given two waves, find phase angle

In summary: In this specific book, the author is a college professor and he or she is trying to teach college level material. It is not meant to be a step by step guide for people who are just starting out in physics.
  • #1
JavierOlivares
14
0

Homework Statement


Given two waves characterized by y1(t) = 3 cos wt; y2(t) = 3 sin(wt + 60deg); does y2(t) lead or lag y1(t) and by what phase angle?

Homework Equations


y1(t) = 3 cos wt; y2(t) = 3 sin(wt + 60deg)
The formula I'm given in my book Fundamentals of Applied Electromagnetics is:
y(x,t) = A cos (2pit/T - 2pi/lamda + reference phase)

The Attempt at a Solution


This question seemed really easy. I figured you just shift the sin wave 60 degrees to the left and I can see by this that y2(t) lags y1(t). I searched everywhere of phase difference and I get the peak:
(distance/wavelength)*2*pi.
The wavelength I assume is arbitrarily 2pi because I'm given w. I easily thought the answer would be 60 degrees but the back of my book says 54 degrees.

I'm completely lost and I've tried multiple things. This seems too easy and I'm frustrated (I need to vent a little). Any help would be greatly appreciated. First time posting a homework problem. I'm on winter break and seems like I should know this in the back of my head. My book is for juniors in college and I'm a little embarrassed for asking this. But I guess for Ch.1 they assume I should easily know this. Ch.1 is very vague and literally goes through my physics class in 1 chapter.
 
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  • #2
Well, trasform cosine in sine or sine in cosine, i suggest using: cosx= sin (x + π/2)

y1 (t) = 3 sin (ωt + π/2)

Then you can compare the two functions to find the phase angle
 
  • #3
If I do what you said I get 30 degrees. Am I missing something?
 
  • #4


I got this formula from watching this video. Φ= (2π/λ)*Δx +ΔΦ0 where Δx = distance traveled from origin of both sources to some point. ΔΦ = inherent phase difference. I changed the first y1 equation into a sine equation. I now have:

y1(t) = 3*sin(ωt +π/2 )
y2(t) = 3*sin(ωt + π/3)

The inherent phase is ΔΦ = Φ2 - Φ1 = π/3-π/2= -π/6 or cause (y1 leads π/6) which is 30°.
λ=2π so Δx = the length it takes for the wave to arrive at a source.
For Δx = I get the same thing = π/3-π/2= -π/6 which adding together gives me 60° phase difference.

In the back of my book I get 54°. I still don't understand what I'm doing wrong. Is there anyone who can help me?
 

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  • #5
The answers in books are wrong quite often.
 

1. What is a phase angle?

A phase angle is a measurement of the fraction of a complete cycle that has elapsed between two waves. It is typically measured in degrees or radians and is used to describe the relationship between the two waves.

2. How do you find the phase angle between two waves?

To find the phase angle between two waves, you must first determine the starting point of each wave. Then, you can calculate the difference in degrees or radians between the two starting points to determine the phase angle.

3. Can the phase angle be negative?

Yes, the phase angle can be negative. This occurs when the second wave lags behind the first wave, resulting in a phase angle that is less than 180 degrees or pi radians.

4. How is the phase angle used in wave interference?

The phase angle is used to describe the relationship between two waves in wave interference. When the phase angle is 0 degrees or 2 pi radians, the waves are said to be in phase and constructive interference occurs. When the phase angle is 180 degrees or pi radians, the waves are said to be out of phase and destructive interference occurs.

5. Can the phase angle change over time?

Yes, the phase angle can change over time. This can occur if the frequency or wavelength of one of the waves changes, causing a shift in the starting points of the waves and thus changing the phase angle. Additionally, the phase angle may change if the waves are traveling through a medium with varying properties, such as temperature or density.

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