# #y(x,t) = Ce^(iomegax//c)e^(-iomegat) = Ce^(ialpha)#, where #alpha = (2pix)/lambda - 2pi nut#... Show that for standing waves, #y(x,t) = Asin((npix)/L)sin(2pin nu_0 t + delta)#? You may want to note the boundary conditions of a vibrating guitar string.

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This is from pp. 26 - 30 in this book:

https://www.dropbox.com/s/fsps97s08o69lp0/Pilar-Ch1.pdf?dl=0

**Going from Eq. (1-86) to (1-87) is where I am mainly stuck.**

#A# is a different constant than #C# , and #delta# is a phase factor.
#nu_0# is the standing wave frequency at the lowest energy level, while the frequency #nu# is dependent on what #n# , the principal quantum number, is. #n nu_0 = nu# .
#lambda# is its wavelength.
#L# is the length of the quantum mechanical box for a given principal quantum number #n# .

Using Euler's formula, and the following relations...

#(2pix)/lambda = (omegax)/c#
#2pinut = 2pin nu_0t = omegat#
#lambda = (2L)/n#

...I got to the point where

#y(x,t) = [Ccos(2pi nut) - iCsin(2pi nu t)]{cos((2pix)/lambda) + isin((2pix)/lambda)]#

#= [Ccos(2pin nu_0t) - iCsin(2pin nu_0t)]{cos((npix)/L) + isin((npix)/L)]# .

How would I get to #Asin((npix)/L)sin(2pin nu_0 t + delta)# ?

This is from pp. 26 - 30 in this book:

https://www.dropbox.com/s/fsps97s08o69lp0/Pilar-Ch1.pdf?dl=0

Going from Eq. (1-86) to (1-87) is where I am mainly stuck.

#A# is a different constant than#C# , and#delta# is a phase factor.#nu_0# is the standing wave frequency at the lowest energy level, while the frequency#nu# is dependent on what#n# , the principal quantum number, is.#n nu_0 = nu# .#lambda# is its wavelength.#L# is the length of the quantum mechanical box for a given principal quantum number#n# .

Using Euler's formula, and the following relations...

#(2pix)/lambda = (omegax)/c# #2pinut = 2pin nu_0t = omegat# #lambda = (2L)/n#

...I got to the point where

#y(x,t) = [Ccos(2pi nut) - iCsin(2pi nu t)]{cos((2pix)/lambda) + isin((2pix)/lambda)]#

#= [Ccos(2pin nu_0t) - iCsin(2pin nu_0t)]{cos((npix)/L) + isin((npix)/L)]# .

How would I get to

##### 3 Answers

I'll partially answer, as a tutorial, the missing parts causing confusion:

**(1-86) to (1-87) - Superimposing Sines and Cosines **

(1-86) :

# g(t) = Acos omega t + Bsin omega t #

(1-87) :# g(t) = Csin(omega t + delta) #

The steps here are a typical "what happens if we assume" mathematical approach. If you graph any function of the form:

#g(t) = Acos omega t + Bsin omega t #

Then it will *appear* to be sinusoidal. so the question is "what happens if we assume" that is is indeed sinusoidal. If so the mathematics should work, otherwise we would get an ambiguity or contradiction. So let us assume that (1-86) can be written in the form (1-86).

That is, we assume that we can write as an identity:

# Acos omega t + Bsin omega t -= Csin(omega t + delta) #

# " " = C(sin omega t cos delta + cos omegat sin delta) #

# " " = Csin delta cos omega t + Ccos delta sin omega t#

As this is an identity we can compare coefficients of

# sin omega t : B = Ccos delta # ..... [A]

# cos omega t : A = Csin delta # ..... [B]

If we take

# A^2 + B^2 = C^2sin^2delta + C^2cos^2 delta #

# A^2 + B^2 = C^2(sin^2delta + cos^2 delta ) #

# A^2 + B^2 = C^2 => C=sqrt(A^2 + B^2) #

And if we take

# (Csin delta)/(Ccosdelta) = A/B => tan delta=A/B#

# :. delta = arctan(A/B) #

And so we can indeed write any trigonometric function of a sum of

Hence we have:

# Acos omega t + Bsin omega t -= Csin(omega t + delta) #

NB: We could equally write [1-86] in the form

# g(t) = C cos(omega t + delta') # and we would find#delta' lt 0#

The choice we make is based on the signs that we get in the sum/difference angle formulas, and if we want the phase shift to be positive or negative, eg:

# sin(A+B)=sinAcosB+cosBsinA rarr Csin(theta + alpha) #

# sin(A-B)=sinAcosB-cosBsinA rarr Csin(theta - alpha) #

# cos(A+B)=cosAcosB-sinAsinB rarr Ccos(theta - alpha) #

# cos(A-B)=cosAcosB+sinAsinB rarr Ccos(theta + alpha) #

{Notes: This technique is typically used to solve any trigonometric equation of the form

See below.

#### Explanation:

To form a standing wave we need a propagation in both sides for the same waveform. So given

To form standing waves we need end conditions for reflection.

With Steve's answer as a supplement, I can provide a full solution to this. However, I'll start from the original solutions to the wave equation instead, which to me makes more sense.

The **wave equation** was:

#(d^2y)/(dx^2) - 1/c^2 (d^2y)/(dt^2) = 0# (with

#c# being the velocity of the wave, not the speed of light.)

to which the solutions were given in the book linked in the description as:

#y(x,t) = f(x)g(t)#

#= [c_1e^(iomegax//c) + c_2e^(-iomegax//c)][a_1e^(iomegat) + a_2e^(-iomegat)]# where:

#omega = 2pinu# is the angular frequency and#nu# is the frequency of the moving wave.#c# in this equation is the speed of light, not the velocity of the wave.

Noting the relationship of the wavelength to the frequency (

#y(x,t) = [c_1e^(i cdot 2pix//lambda) + c_2e^(-i cdot 2pix//lambda)][a_1e^(i cdot 2pinu t) + a_2e^(-i cdot 2pinu t)]#

Using **Euler's relation**

#ul(f(x)) = c_1(cos(2pix//lambda) + isin(2pix//lambda)) + c_2(cos(2pix//lambda) - isin(2pix//lambda))#

#= ul(Bcos(2pix//lambda) + Asin(2pix//lambda))# for which we define

#B = c_1 + c_2# and#A = ic_1 - ic_2# .

Then, we can rewrite

#ul(g(t) = A_1cos(2pinu t) + A_2sin(2pi nu t))# where

#A_1 = a_1 + a_2# and#A_2 = ia_1 - ia_2# .

As Steve has gotten, we could rewrite

#ul(g(t) = Csin(2pinu t + delta))# ,if

#C = sqrt(A_1^2 + A_2^2)# and#delta = arctan(A_1/A_2)# .

As this wave function would *later be re-normalized* to find the constant out in front, we can ignore the

#y(x,t) = [Bcos(2pix//lambda) + Asin(2pix//lambda)]sin(2pi nu t + delta)#

Now, note the **boundary conditions** of the standing wave in a box of length

#f(x=0) = f(x=L) = 0#

#=> Bcos(2piL//lambda) + Asin(2piL//lambda) = Bcos(0) + Asin(0) = 0#

The right-hand side implies that

#(2piL)/lambda = npi#

Thus,

#y(x,t) = Asin(2pix//lambda)sin(2pi nu t + delta)#

#= Asin((2pinx)/(2L))sin(2pi nu t + delta)#

#= Asin((npix)/(L))sin(2pi nu t + delta)#

One last order of business is to consider that for a **standing wave**, modeled by a so-called particle in a box system with energy levels quantized by the principal quantum number

The wavelength was found earlier to be

#nu = c cdot n/(2L)#

If we then realize that if

#nu_0 = c/(2L)# (or if we realize that

#lambda = 2L# for#n = 1# )

Then, we get **final solution** is:

#color(blue)(y(x,t) = Asin((npix)/(L))sin(2pi n nu_0 t + delta))#