Answer to Question #172066 in Differential Equations for Albert

Question

The point of trisection of a string are pulled aside through a distance ∈ on opposite sides of the position of equilibrium and the string is released from rest. Derive an expression for the displacement of the string at any subsequent time and show that the midpoint of the string always remains at rest..

Answer 1

We know that the wave equation is $\frac{\partial^2 y}{\partial t^2} =c^2 \frac{\partial^2 y}{\partial x^2} \\$

Given the boundary condition are as follows

(i)y(0, t)=0, t>0

(ii)y(l, t)=0, t>0

$(iii)\frac{\partial y}{\partial t}(x, 0)=0, 0<x<l$

Let OA=l, where 0(0, t) and A(l, t).

Let B($\frac{l}{3}, h)$ and C($\frac{2l}{3}, h)$ be the point of the string.

The line equation OB is given by,

$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

Here,

$(x_1, y_1)=(0, t), (x_2, y_2)=({\frac{l}{3}, h})\\ \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \implies \frac{y-t}{h-t}=\frac{x-0}{\frac{l}{3}-0}\\ y(x, t)=\frac{3x}{l}(h-t)+t,\\ Put, t=0\\ y(x, 0)={3xh}{l}, 0<x<\frac{l}{3}\\ \text{The equation of the line BC is}, \\ \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \\ (x_1, y_1)=(\frac{l}{3}, h), (x_2, y_2)=({\frac{2l}{3}, -h})\\ \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} , gives, \\ y(x, t)=\frac{3h(l-2x)}{l}\\ Put, t=0\\ y(x, 0)=\frac{3h(l-2x)}{l}, \frac{l}{3}<x<\frac{2l}{3}\\, \text{we solve for CA too}\\ \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \\ (x_1, y_1)=(\frac{2l}{3}, -h), (x_2, y_2)=(l, t)\\ \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \implies \\ y+h=\frac{3x-2l}{l}(t+h)\\ y(x, t)=\frac{3x-2l)}{l}(t+h)-h\\ Put, t=0\\ y(x, 0)=\frac{3h(x-l)}{l}, \frac{2l}{3}<x<l\\$ ,

Hence we have the boundary condition solved.

We then know that the proper complete solution of the wave is,

y(x, t)=(Acospx+Bsinpx)(Ccoscpt+Dsincpt),

Using the boundaries to solve for all cases, we see that there is no displacement at the midpoint x=$\frac{1}{2}$ on the string.(i.e) the midpoint is at rest