Question

Determine the deflection u (x, t) of a vibrating string of length L, which has its ends fixed,
corresponding to a zero initial velocity and an initial deflection given by the function
f(x)=A(x) for 0<=X<=L/4
=AL/4 for L/4<=X<=3L/4
=A(L-X) for 3L/4<=X<=L

Answer 1

Answer on Question #38893, Math, Other

The answer to given problem might be obtained using Fourier method. The formula is

$$u(x, t) = \sum_{n=1}^{\infty} \left(A_n \cos \pi n \frac{v}{L} x + B_n \sin \pi n \frac{v}{L} t\right) \sin \pi \frac{n}{L} x.$$

Using initial conditions,

$$u(x, 0) = \varphi(x), u_t(x, 0) = \psi(x) \quad \text{it is easy to obtain Fourier coefficients} \quad A_n = \frac{2}{L} \int_0^L \varphi(x) \sin \pi \frac{n}{L} x \, dx$$ $$B_n = \frac{2}{\pi n v} \int_0^L \psi(x) \sin \pi \frac{n}{L} x \, dx.$$

For current problem, $\psi(x) = 0$. Thus,

$$A_n = \frac{2}{L} \left( A \int_0^{\frac{L}{4}} x \sin \pi \frac{n}{L} x \, dx + \frac{A L}{4} \int_{\frac{L}{4}}^{\frac{3L}{4}} \sin \pi \frac{n}{L} x \, dx + A \int_{\frac{3L}{4}}^{\frac{L}{4}} (L - x) \sin \pi \frac{n}{L} x \, dx \right) =$$ $$\frac{2}{L} \left( \frac{A L}{n^2 \pi^2} \left( \frac{-1}{4} L n \pi \cos \pi \frac{n}{4} + L \sin \pi \frac{n}{4} \right) + \frac{A L}{4} \frac{2 L \sin \pi \frac{n}{4} \sin \pi \frac{n}{2}}{n \pi} + A L^2 \frac{(n \pi \cos 3 \frac{\pi}{4} n + 4 \sin 3 \frac{\pi}{4} n)}{4 n^2 \pi^2} \right) =$$ $$= \frac{2}{L} \left( \frac{A L^2}{n^2 \pi^2} \left( 2 \sin \pi \frac{n}{2} \cos \pi \frac{n}{4} \right) \right) = \frac{4 A L}{\pi^2 n^2} \sin \pi \frac{n}{2} \cos \pi \frac{n}{4}.$$

The solution is $u(x,t) = \sum_{n=1}^{\infty} \frac{4A L}{\pi^2 n^2} \sin \pi \frac{n}{2} \cos \pi \frac{n}{4} \sin \pi \frac{n}{L} x$.