User:Egm6322.s09.bit.sahin/r5

Treat tension $$\tau \left (x,t \right )$$ and slope angle $$\theta\left (x,t \right ) $$ which are both algebraic quantities. According to the following figure we can write that:

$$\Sigma F_{x}=\tau \left (x \right )cos\theta \left (x \right )+\tau \left (x+dx \right )cos\theta \left (x+dx \right )=0$$ (1)

$$\Sigma F_{y}=\tau \left (x \right )sin\theta \left (x \right )+\tau \left (x+dx \right )sin\theta \left (x+dx \right )+p\left (x \right)dx-mdxw_{tt}=0$$ (2)

For equation 1, we can assume $$\theta$$ is small, making $$cos (\theta) \cong 1$$, leading to:

$$\tau \left (x+dx \right )=-\tau \left (x \right )=\tau =constant$$ for all x. Therefore,

$$\tau \left (x+dx \right )=\tau$$

$$\tau \left (x \right )=-\tau$$

For equation 2:

$$\Sigma F_{y}=-\tau sin\theta \left ( x\right )+\tau sin\theta \left (x+dx \right )+p\left (x \right )dx-mdxw_{tt}=0$$

For small $$\theta$$,

$$\Sigma F_{y}=0=\tau \frac{\partial \theta }{\partial x}dx+h.o.t+p\left (x \right )dx-mdxw_{tt}$$

Eventually, we obtain the equation of motion for membrane as:

$$\tau w_{xx}+p=mw_{tt}$$

Part2
$$\frac {1}{rR} \frac {\partial} {\partial r} (r \frac {dR} {dr})$$ $$+ \frac {1}{r^2 \Theta} \frac {d^2 \Theta}{d \theta^2 \theta}$$ $$- \frac {1}{T} \frac {dT}{dt}=0$$

Since $$\left (r,\theta ,t \right )$$ are independent varibles,

$$f\left (r,\theta \right )+g\left (t \right )=0$$  (1)

$$f\left (r,\theta \right )=-g\left (t \right )=\lambda =constant$$   (2)

Therefore,

$$\frac{1}{T}\frac{dT}{dt}=-\lambda$$

$$\frac{dT}{T}=-\lambda dt$$

$$logT=-\lambda t+k_{1}$$

$$T\left (t \right )=exp\left (-\lambda t+K_{1} \right )$$

$$T\left (t \right )=exp\left (-\lambda t \right )expK_{1}$$

Defining $$K_{2}=expK_{1}$$, we have

$$T\left (t \right )= K_{2}exp\left (-\lambda t \right )$$

It is easily seen that as $$t\rightarrow +\infty $$, $$exp\left (-\lambda t \right )$$ approaching to zero. Therefore we choose $$\lambda >0$$ at the beginning.