#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.

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# and #cos omega t# to get:

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

If we take #[A]^2 + [B]^2#, and use the identity #cos^2theta+sin^theta -= 1# then:

# 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 #Eq[B] div Eq[A]# we get:

# (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 #sin# and #cos# of some common frequency #omega# as a single sinusoidal function of the same frequency #omega# with a phase shift #delta# and amplitude #C#

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 #Acos theta + Bsin theta = c#, as that equation can be re-written inb the form #Rsin(theta+alpha)=c# or #Rcos(theta+alpha)=c#, which are then trivial to solve].

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

#y(x,t)=y_0 sin(mu x + omega t)# the standing wave occurs when

#y_s(x,y)=y_0 sin(mu x + omega t)+y_0 sin(mu x - omega t)# and simplifying

#y_s(x,y)=y_0 sin(mu x)sin(omegat)#

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 (#lambda = c/nu#, with #omega/(2pi) = nu#), this can also be rewritten as:

#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 #e^(pmix) = cosx pm isinx#, one can first rewrite #f(x)# in terms of #sin# and #cos#:

#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 #g(t)# in terms of #sin# and #cos# to similarly get:

#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 #C# and so far have:

#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 #L# fixed at #0# and #L#:

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

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

The right-hand side implies that #B = 0#, so in order for the left-hand side to equal #0#, we must have:

#(2piL)/lambda = npi#

Thus, #lambda = (2piL)/(npi) = (2L)/n#, and we can rewrite #y(x,t)# once more as:

#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 #n#:

The wavelength was found earlier to be #lambda = (2L)/n#, so since #nu = c/lambda#, we have:

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

If we then realize that if #n = 1#, we can define the frequency at #n = 1#, the ground state, as:

#nu_0 = c/(2L)#

(or if we realize that #lambda = 2L# for #n = 1#)

Then, we get #nu = n nu_0#. Therefore, our final solution is:

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