Deriving Wave Function for One-Dimensional Sinusoidal Wave

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SUMMARY

The discussion centers on deriving the wave function for a one-dimensional sinusoidal wave, specifically the equation y = sin(2πx/λ). Participants clarify that the function's period is λ, which is achieved by adjusting the sine function's argument to 2πx/λ. The confusion arises from the reference to Figure 16.10, which depicts sin(θ) but is not directly related to the derivation of the wave function. The key takeaway is that the periodicity of the function is determined by the wavelength λ, not the graph itself.

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Homework Statement
Please see below and https://openstax.org/books/university-physics-volume-1/pages/16-2-mathematics-of-waves for more details.
Relevant Equations
Please see below
Where did they get the equation in circled in red from? It does not seem that it can be derived from the graph below.
1672944351154.png

Many thanks
 
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Figure 16.10 is just a graph of \sin(theta). It is not referenced in the extract you have posted; why then you would expect the highlighted part to be directly derivable from a figure which is nowhere referred to? Perhaps look at the part of the text where the figure is actually referenced.

I suspect that, prior to the extract you have posted, it is stated that y should have period \lambda. Sine, of course, has a period of 2\pi, so to get a function with period \lambda you have to use \sin (2\pi x/\lambda), so that \theta = 2\pi x/\lambda is equal to 2\pi when x = \lambda.
 
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pasmith said:
Figure 16.10 is just a graph of \sin(theta). It is not referenced in the extract you have posted; why then you would expect the highlighted part to be directly derivable from a figure which is nowhere referred to? Perhaps look at the part of the text where the figure is actually referenced.

I suspect that, prior to the extract you have posted, it is stated that y should have period \lambda. Sine, of course, has a period of 2\pi, so to get a function with period \lambda you have to use \sin (2\pi x/\lambda), so that \theta = 2\pi x/\lambda is equal to 2\pi when x = \lambda.
Thanks for your reply @pasmith! You meant to say that "it is stated that x should have a period of λ" instead of "
1672949737419.png
" correct?

I guess they took an arbitrary point along the wave to for ratio of phase to wavelength which they also could of picked a point on the wave which has a π phase which has a wavelength λ/2.
1672950703818.png

Many thanks
 
Callumnc1 said:
You meant to say that "it is stated that x should have a period of λ" instead of "
1672949737419-png.png
" correct?
incorrect. In ##y=\sin(2\pi x/\lambda)##, it is the value of y that repeats as x increases by λ, so we say y has period λ.
 
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haruspex said:
incorrect. In ##y=\sin(2\pi x/\lambda)##, it is the value of y that repeats as x increases by λ, so we say y has period λ.
Thank you @haruspex !
 

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