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Last modified: November 18, 2021 ✓

One Dimensional Wave Equation
-----------------------------

This equation applies to both longitudinal and 
transverse waves.  In the case of transverse waves 
this is easy to see.  In the case of longitudinal 
waves, particles in the area where the fluid is 
compressed or expanded can be shown to move back 
and forth in a sine wave over time.  This is also 
apparent when studying the simple harmonic motion 
of a spring.  Mathematically we can describe waves 
in a way that is independent of the medium by 
expressing the physical quantity that is "waving" 
as a function of position, Q(x,t), and time.  In a 
sound wave, Q could be pressure or density or 
longitudinal displacement.  In a light wave, Q can 
be any of the components of the electromagnetic 
field, or any of the components of the EM potential. 

Consider a small segment of rope under tension.



T1cosα = T2cosβ = T  
                                                     
T1sinβ - T2sinα = (ρΔx)∂2y/∂t2 

Where the RHS is equivalent to ma (ρ = mass per 
unit length).

T1sinβ/T1cosβ - T2sinα/T2cosα = (ρΔx)∂2y/∂t2 

tanβ - tanα = (ρΔx)∂2y/∂t2 

tanα = ∂y/∂x|x   tanα = ∂y/∂x|x+Δx  

(1/Δx){∂y/∂x|x - ∂y/∂x|x+Δx} = ∂y/∂x|x+Δx

∂2y/∂x2 = (ρ/T)∂2y/∂t2

∂2y/∂x2 = (1/v2)∂2y/∂t2  v is phase velocity

Alternative proof:

Consider an array of masses, m, separated by a
distance h and connected by masslesss springs each 
with a spring constant, k



u(x) refers to the displacement of the mass at x.  
The forces exerted on the mass at x + h are:

From Newton's Law:

F = ma = m∂2u(x + h)/∂t2

From Hooke's Law:

For a displacement to the right  spring B is 
compressed and spring A is stretched.  Thus,

FB = k[u(x + 2h) - u(x + h)]    

FA = -k[u(x + h) - u(x)]

F = FA + FB

  = k[u(x + 2h) - u(x + h)] + (-k[u(x + h) - u(x)])  

m∂2u(x + h)/∂t2 = k[u(x + 2h) - u(x + h)] 

                  + (-k[u(x + h) - u(x)])

∂2u(x + h)/∂t2 = (k/m)k[u(x + 2h) - u(x + h)] 

                 - k[u(x + h) - u(x)])

If the array of weights consists of N weights spaced 
evenly over the length, then L = Nh, the total mass, 
M = Nm, and the total spring constant of the array, 
K = k/N.  Then,

k/m = KN/(M/N) = KN2/M = KL2/h2M

and we can write:

∂2u(x + h)/∂t2 = (KL2/M)[k[u(x + 2h) - u(x + h)] 

                 - k[u(x + h) - u(x)])/h2

As N -> ∞,  h -> 0 the quantity, [k[u(x + 2h) - u(x + h)] 
- k[u(x + h) - u(x)]/h2, becomes the second derivative 
of u(x) w.r.t. x.  Thus,

∂2u(x)/∂t2 = (KL2/M)∂2u(x)/∂x2

Now, [K] = N/m or Kg/s2 ∴ [(KL2/M)] = [v2]

Therefore, we can write:

∂2u(x)/∂t2 = v2∂2u(x)/∂x2

Solutions
---------

Traveling wave:  

y = Asin(2π/λ)(x +/- vt)

  = Asin(kx - ωt)
                     l->r            r->l
Standing wave:  y = Asin(kx - ωt) + Asin(kx + ωt) 

                  = 2Asinkxcosωt   

the second term is the reflected wave (180° phase 
shift).  For a string of length, L, fixed at both 
ends:

nλ/2 = L, k = 2π/λ => k = nπ/L

Therefore,

y = 2Asin(nπx/L)cosωt   

∂2y/∂x2 = -(2An2π2/L2)sin(nπx/L)cosωt

(ρ/T)∂2y/∂t2 = -(2Aρ/T)sin(nπx/L)cosωt

=> ω = (nπ/L)√(T/ρ)

     = vnπ/L

=> f = nv/2L

 

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