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Posted (edited)

The function

 

φ(x) = a(sin 2πx/λ) + b(cos 2πx/λ)

 

represents the superposition of two harmonic waves with the same wavelength lambda. Show that phi is also harmonic with the same wavelength, and can be written as

 

φ(x) = A sin((2πx/λ) + α)

 

where A = √(a² + b²)

 

the best I can come up with is

 

φ(x) = A sin ((2πx/λ) + α)

= (√( a² + b²)) (sin(2πx/λ) cos α + cos(2πx/λ) sin α)

Edited by buttacup
  • 11 months later...
Posted (edited)

I never did figure this one out...........maybe if I format the question properly someone can throw me a clue......

 

The function

 

[math] \phi (x) = a(sin \frac {2 \pi x}{\lambda}) + b(cos \frac {2 \pi x}{\lambda}) [/math]

 

represents the superposition of two harmonic waves with the same wavelength [math] \lambda [/math]. Show that [math] \phi [/math] is also harmonic with the same wavelength, and can be written as

 

[math]\phi (x) = A sin(\frac{2\pi x}{\lambda})+\alpha[/math]

 

[math]A = \sqrt{a^2 + b^2}[/math]

 

the best I can come up with is

 

[math]\phi (x) = A sin ( \frac{2\pi x}{\lambda} + \alpha) = (\sqrt{a^2 + b^2})( cos \alpha (sin\frac{2 \pi x}{\lambda}) + sin \alpha (cos\frac{2\pi x}{\lambda}))[/math]

 

how would

 

[math]a=cos \alpha \sqrt{a^2 + b^2}[/math]

 

and

 

[math]b=sin \alpha \sqrt{a^2 + b^2}[/math]

 

something tells me this is really easy.......

 

........thread titles............

Edited by buttacup
Posted

As always in such questions I'd use the relation to the complex exponential function [math]e^{ix} = \cos x + i \sin x[/math]. From that it follows that

 

[math] \sin x = \frac{ e^{ix} - e^{-ix} }{2i} [/math]

[math] \cos x = \frac{ e^{ix} + e^{-ix} }{2} [/math]

 

Rewriting [math] A \sin cx + B \cos cx [/math] that way and realizing that appearing terms like [math]( B + iA )[/math] can, like any other complex number, be written as a magnitude times a complex phase [math]( B + iA ) = \sqrt{A^2 + B^2} \exp ( i \underbrace{ \tan^{-1} \frac AB}_{=:\delta} ) [/math] you pretty much have the result.

 

That way of solving the question assumes you know the relation of exp(ix) to the trigonometric functions, know complex numbers and know the exponential function. But on the plus side you don't have to remember or look up all the trig identities.

Posted
But on the plus side you don't have to remember or look up all the trig identities.

 

It was on the list..............remembering that is!:P

 

Thx for the insight.

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