Question

Transverse waves on a string obey the equation -where u(x,t) is the time Cu (a) Ifat t=0, the transverse displacement u(x,0) =_and__ = 0 find u(x,t)Use the 1+x Cx general solution of the wave equation. (b) Ifthe two ends ofthe string are fixed at x=0 and x=a and at t=0 u(x,0)-0, adx,t= 0) = dx(a-x) , use separation of variables to obtain u(x,t). Cx

Answer 1

We have the differential equation for the transverse wave is

d²u/dx²=(1/v²)d²u/dt². d is the partial differential operator

We can write above v equation as

X^"t=(1/v²)(t^"x).

x^"/x=t^"/t=v².

Finally two equations become

X^"-v²X=0. And t^"-v²t=0.

And solutions become from equations D²-v²=0,roots D=+v,-v. And general solution become

u(x,t)=(c₁e^vx+c₂e^-vx)(c₃e^vt+c₄e^-vt). Transverse displacement.

Where x is the position,v is the wave speed and t is the time.

(a). Given initial conditions are. u(x,0)=(1/(1+x²)). And du/dx=0.

u(x,0)=(c₁e^vx+c₂e^-vx)(c₃+c₄)=(1/(1+x²))-->c₃+c₄=(1/(1+x²))/(2c₁e^vx)

du/dx=(c₁ve^vx-c₂ve^-vx)(c₃e^vt+c₄e^-vt)=0.

That means

V(c₁e^vx-c₂e^-vx)=0

C₁e^vx=c₂e^-vx.

C₂=c₁e^2vx.

Finally general solution become if we are taking coefficients as zeros

u(x,t)=2e^vx(1/(1+x²)/(2e^xv)=(1/(1+x²)).

(b).Now,given initial conditions are u(x,0)=0. And substitute in the general solution

u(x,0)=c₃+c₄=0 an c₃=-c₄.

u(0,t), u(a,t) are fixed.means u(0,t)=0 and u(a,t)=0. Fixed means displacement will become zero.

u(0,t)=(c₁+c₂)=0 and c₁=-c₂.

And u(a,t)=c₁e^ax+c₂e^-ax=0.c₂=-c₁e^2ax.

du(x,t=0)/dx=Ax(a-x)=(c₁e^vx-c₂e^-vx)(c₃+c₄).

So,Final soltiuon by seperating x and t variables is

u(x,t)=-e^2ax(Ax(a-x))=-e^2at(At(a-t))