Question

A string of length 10 𝑓𝑡. is raised at the middle to a distance of 1 𝑓𝑡, and then released. Describe the motion of the string assuming 𝑐 = 1𝑓𝑡. 𝑠 −1. 

Answer 1

Equation of AO

$y-0= \frac{x}{5}\\ y= \frac{x}{5}$

Equation of AB

$y= \frac{x-10}{5-10}\\ y= \frac{10-x}{5}\\$

Hence the motion distributed by

$y_{tt}=y_{xx}\\ y_{x,t}=X_{x}*xT''\\ \frac{T''}{T}=\frac{X''}{X}= - λ^2\\ BVP\\ X'' + λ^2X=0\\ X(0)=X(10)=0; T''+λ^2T=0; T'(0)=0\\ X(x)= c_1 \cos λ x + c_2 \sin λ x; T(t)= c_3 \cos λ t + c_4 \sin λ t\\ For \space X(0)=0 \implies c_1=0, T'(0)=0 \implies c_4 =0\\ X (10)=0 \implies 10 λ= n \pi \implies λ = \frac{n \pi}{10} ; T_n= \cos (\frac{n \pi}{10}t)\\ X_n= \sin(\frac{n \pi}{10}x)\\ Y(x,t)=\sum_{n=1}^{\infin} B_n\sin(\frac{n \pi}{10}x)* \cos (\frac{n \pi}{10}t)\\ Y(x,0)= f(x)=\sum_{n=1}^{\infin} B_n\sin(\frac{n \pi}{10}x)\\ B_n=\frac{2}{10} \int_0^{10} f(x) \sin(\frac{n \pi}{10}x) dx\\ = \frac{1}{5}[ \int_0^{5} \frac{1}{5} x \sin(\frac{n \pi}{10}x) dx+ \int_5^{10} \frac{1}{5}(10- x) \sin(\frac{n \pi}{10}x) dx]\\ = \frac{8}{n^2 \pi^2}\sin(\frac{n \pi}{2})\\ \therefore Y(x,t)=\sum_{n=1}^{\infin} \frac{8}{n^2 \pi^2}\sin(\frac{n \pi}{2}) \sin(\frac{n \pi}{10}x)* \cos (\frac{n \pi}{10}t)\\$