Question 9.2: A string that is fixed at both ends is initially deformed by...
A string that is fixed at both ends is initially deformed by displacing its middle by a distance d before releasing it without an initial velocity (Figure 9.5). a) Write down the expression of the produced wave. b) Calculate the total energy of the excited modes and verify that it is equal to the work done in displacing the middle of the string by a distance d .
a) The initial conditions (for t = 0 ) are in this case \dot{u} (x,0)=0 and
u(x,0)=2xd/L \ \ \ (\text {for} \ 0\lt x\lt L/2) \ \ \ \text {and} \ \ \ u(x,0)=2d(1-x/L) \ (\text {for } L/2\lt x\lt L).The expressions of [9.34] give the Fourier coefficients
B_n=\frac{2}{L} \int_{0}^{L}{dx} \ \ u(x,0)\sin(\frac{n\pi x}{L} ), \ \ \ \ C_n=\frac{2}{n\pi v} \int_{0}^{L}{dx} \ \ \dot{u} (x,0) \sin(\frac{n\pi x}{L} ). [9.34]
B_n=(-1)^p8d/\pi ^2(2p+1)^2 \ \ \ \ \text {if }n=2p+1 \ \ \ \ \text {and} \ \ \ B_n=0 \ \ \ \text {if } \ \ n=2p \ \ \ \text {and} \ \ \ C_n=0.Thus, the wave can be written as
u=\frac{8d}{\pi ^2} \Sigma _{P\geq 0} \frac{(-1)^p}{(2p+1)^2} \cos\left[(2p+1)\pi v\frac{t}{L} \right] \sin\left[(2p+1)\pi \frac{x}{L} \right] .We note that the amplitude of the modes decreases like 1/(2p+1)^2 , this means that only the first few modes are effectively excited.
b) The total energy of the string is given by expression [9.37], i.e.
U=\int_{0}^{L}{dx} \ \ U_l (x)=¼ Lm_l\Sigma _{n} (\omega _nA_{n)} \ ^2 =\Sigma _{n} \ U_n. [9.37]
U=\frac{1}{4} Lm_l \Sigma _{n}(\omega _nA_{n)} \ ^2=\frac{16d^2}{\pi ^4} \Sigma _{P} \frac{1}{(2p+1)^4} m_l\omega _{2P+1} \ ^2=\frac{16d^2F}{L\pi ^2} \Sigma _{P} \frac{1}{(2p+1)^2} .The sum \Sigma (2p+1)^{-2} is equal to π²/8; thus, the total energy is 2d^2F/L . The initial
total energy is the potential energy. This is the work done to lengthen the string from L to 2\sqrt{L^2/4+d^2} , i.e. F\left[\sqrt{L^2/4+d^2}-L \right] \approx 2Fd^2/L. Thus, the total energy of the standing wave on the string is equal to the work done to excite the string.