Question

Example. A stretched string of length 50cm is set oscillating by displacing its mid distance of 2cm from its rest point a equation-=- where c21, to determine the resulting motion u(x, t). 2 2 iKnx) 25 50 x(cm)

Answer 1

Here,length of the string is 50cm after stretching it and given wave equation is

(d²u/dx²)=(1/c²)(d²u/dt²). Where d is a partial differential operator

We can also write this equation as

X^"t=(1/c²)T^"x.

x^"/x=t^"/t=c².

And substitute the given value of c² in the above differential equation

This can be written as

x^"/x=t^"/t=1.

And by separation of variables ODE equations become

x^"-x=0. and t^"-t=0.

We can simplify above equations as

(D²-1)x=0 and (D²-1)t=0.

As D=+-1 for both x and t,General solution will become

u(x,t)=(c₁e^x+c₂e^-x)(c₃e^t+c₄e^-t).

Initial Conditions from the graph

u(0,t)=0 and u(x,0)=0

u(50,t)=(2/25)(e²⁵⁽²⁾-e-^25(-2))=(2/25)(e⁵⁰-e^-50).It is triangular wave.

u(0,t)=c₁+c₂=0 and u(x,0)=c₃+c₄=0.

u(50,t)=c₁e⁵⁰+c₂e^-50=(2/25)(e⁵⁰-e^-50).and we got c₁=-c₂.

After substituting above conditions

We will get c₁=-(2/25) and c₂=(2/25).

u(2,x)=f(x)=(2/25)(2-25)=(2/25)(-23)=(c₃e²+c₄e^-2) and c₃=-c₄

and c₃=(2/25)(-23)=-46/25.

Substitute above values and we get general solution as

u(x,t)=2/25(e^-50-e⁵⁰)(-46/25)(e^-2-e²).