Question 8.1: Consider a string of density σ that is stretched between two...

(a) The part \overline{PQ} of the string has the mass σΔx and the velocity ∂y/∂t .
Its kinetic energy is then

ΔT = \frac{1}{2} σ Δx \left(\frac{∂y}{∂t}\right)^{2}.                                   (8.24)

The total kinetic energy of the string between x = a and b is

T = \frac{1}{2} σ  \int\limits_{a}^{b}{ \left(\frac{∂y}{∂t}\right)^{2} dx}.                                     (8.25)

The work which is needed to elongate the string from Δx   to   Δl is

dP = T_{0}(Δl −Δx),                        \frac{Δl}{Δx}∼ 1.                                  (8.26)

For small displacements, we have

Δl = (Δx^{2} + Δy^{2})^{1/2} = Δx \left[1+ \left(\frac{∂y}{∂x}\right)^{2} \right]^{1/2} \simeq Δx \left[1+ \frac{1}{2}  \left(\frac{∂y}{∂x}\right)^{2}\right].                            (8.27)

The potential energy for the region x = a   to   x = b is then

P = \frac{1}{2} T_{0} \int\limits_{a}^{b}{\left(\frac{∂y}{∂x}\right)^{2} dx}.                             (8.28)

For a wave y = F(x − ct) propagating in a direction, we have

T = P = \frac{1}{2} T_{0} \int\limits_{a}^{b}{[F^{′}(x −ct)]^{2}  dx},                      c_{2} = \frac{T_{0}}{σ}.                                         (8.29)

Hence, the kinetic and potential energy are equal. If a, b are fixed points, then T and P vary with time. But if we admit that a and b can propagate with the sound velocity c, so that

a = A +ct       and         b = B +ct,                                (8.30)

then P and T are constant:

T = P = \frac{1}{2} T_{0} \int\limits_{A}^{B}{(F^{′}(x))^{2}  dx}.                                     (8.31)

(b)

⇒ \left(\frac{∂y}{∂t}\right)^{2} = C^{2}  sin^{2} \left(\frac{ω}{c} x −ωt\right) ω^{2}.                             (8.32)

Insertion into (8.25) yields (a = 0, b = λ)

T = \frac{1}{2} \frac{T_{0}}{c^{2}}  C^{2} ω^{2} \int\limits_{0}^{λ}{sin^{2} \left(\frac{ω}{c} x −ωt\right) dx} =\frac{1}{2} \frac{T_{0}}{c^{2}} C^{2}ω^{2} · I.                         (8.33)

With the substitution z = (ω/c)x −ωt for the integral I , we find

I = \frac{c}{ω} \int\limits_{−ωt}^{(ω/c)λ−ωt}{sin^{2}   zdz} = \frac{c}{ω} \int\limits_{0}^{(ω/c)λ}{sin^{2}  zdz} = \frac{c}{ω} \int\limits_{0}^{2π}{sin^{2}  zdz}                        (8.34)

⇒ T = \frac{1}{2} \frac{T_{0}}{c^{2}}  C^{2}ω^{2}  \frac{c}{ω} π = \frac{π^{2}C^{2}T_{0}}{λ},                    λ= 2π \frac{c}{ω}.                                    (8.35)

One gets the same expression for the potential energy. Insertion of the numerical values yields