Given two waves, find phase angle

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The discussion revolves around determining the phase relationship between two waves, y1(t) = 3 cos(wt) and y2(t) = 3 sin(wt + 60°). The initial assumption is that y2(t) lags y1(t) by 60 degrees, but the textbook indicates a phase difference of 54 degrees, causing confusion. Participants suggest transforming the cosine function into a sine function to compare the two waves more easily, leading to calculations that yield a phase difference of 30 degrees. Despite multiple attempts and using various formulas, the correct phase angle remains elusive, highlighting the complexity of phase relationships in wave mechanics. The discussion emphasizes the challenges faced when reconciling textbook answers with personal calculations.
JavierOlivares
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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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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
 
If I do what you said I get 30 degrees. Am I missing something?
 


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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The answers in books are wrong quite often.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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