Question about travelling wave equation

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The discussion revolves around a multiple-choice question regarding the traveling wave equation, where the user calculates the wave parameters and arrives at a specific equation. The user questions whether the negative sign in their equation can be disregarded, as the book provides a different answer. Responses clarify that the negative sign cannot simply be ignored, but it represents a phase shift of π, which is mathematically valid. Additionally, the concept of amplitude is discussed, emphasizing that amplitude itself cannot be negative, and the term "magnitude" is suggested as a more accurate descriptor. The conversation highlights the importance of understanding phase shifts and the correct interpretation of wave equations.
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There is a multiple choices question about traveling wave in my book.
Based on the graphic, if T = 2s, the wave equation is ...

fis.jpg


My answer :
ω = 2π/T = 2π/2 = π
k = 2π/λ = 2π/4 = 0,5π → in my country, we use comma (,) for point (.)
y = ±A sin (ωt - kx)
y = -0,5 sin (πt - 0,5πx)
y = -0,5 sin π(t - 0,5x)

but the answer on my book is B. Could we throw the negative anyway?
 
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Yoseph Santoso said:
There is a multiple choices question about traveling wave in my book.
Based on the graphic, if T = 2s, the wave equation is ...

View attachment 258146

My answer :
ω = 2π/T = 2π/2 = π
k = 2π/λ = 2π/4 = 0,5π → in my country, we use comma (,) for point (.)
y = ±A sin (ωt - kx)
y = -0,5 sin (πt - 0,5πx)
y = -0,5 sin π(t - 0,5x)

but the answer on my book is B. Could we throw the negative anyway?
You can't just throw away the negative, but you can see it's the same equation as B with an additional phase of ##\pi##.
 
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How do you get the minus sign?
 
PeroK said:
You can't just throw away the negative, but you can see it's the same equation as B with an additional phase of ##\pi##.
Amplitude itself is just distance
It can't be negative.
 
Adityameowmeow said:
Amplitude itself is just distance
It can't be negative.
Nobody was making the amplitude negative. If the equation is ##y=-A\sin(\omega t)##, where A>0, then the amplitude is A. Changing it to ##y=A\sin(\omega t)## would be wrong, but if you wish you can change it to ##y=A\sin(\omega t+\pi)##.
 
Adityameowmeow said:
Amplitude itself is just distance
A better term in English is "magnitude". Just like a vector has "magnitude" and "direction". :smile:
 
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