Question #286502

A taut string of length 2l , fastened at both ends, is disturbed from its position of equilibrium by imparting to each of its points on an initial velocity of magnitude k(2lx-x^2). Find the displacement function y(x,t)

1
Expert's answer
2022-01-12T08:00:29-0500

The displacement y(x,t)y(x,t) of any point 'xx ' of the string at any time 'tt ' is given by

2yt2=a22yx2\frac{\partial²y}{\partial t²}=a²\frac{\partial²y}{\partial x^{2}} (1)---(1)

We have to solve equation (1) satisfying the following boundary conditions.

y(0,t)=0,y(0,t)=0, forfor t0t≥0 (2)----(2)

y(2l,t)=0,y(2l,t)=0, forfor t0t≥0 (3)----(3)

y(x,0)=0,y(x,0)=0, forfor 0x2l0≤x≤2l (4)----(4)

δyδt(x,0)=k(2lxx2),\frac{\delta y}{\delta t}(x,0)=k(2lx-x²), forfor 0x2l0≤x≤2l (5)----(5)

The suitable solution of Eq (1), consistent with the vibration of the string,is

y(x,t)=(Acospx+Bsinpx)y(x,t)=(A\cos px + B\sin px)

(Ccospat+Dsinpat)(C\cos pat+D\sin pat) (6)----(6)

Using boundary conditions (2) in (6), we have

A(Ccospat+Dsinpat)=0A(C\cos pat+D\sin pat)=0 forfor allall t0t≥0

A=0A=0

Using boundary conditions (3) in (6), we have

Bsin2lp(Ccospat+Dsinpat)=0B\sin 2lp(C\cos pat+D\sin pat)=0

forfor allall t0t≥0

Either B=0B=0 or sin2lp=0\sin 2lp=0


If we assume that B=0,B=0, we get a trivial solution.

sin2lp=0\sin 2lp=0

2lp=nπ2lp=nπ

p=nπ2lp=\frac{nπ}{2l}

Where n=0,1,2,...n=0,1,2,...\infty

Using boundary conditions (4) in (6), we have

Bsinpx.C=0B\sin px.C=0

forfor 0x2l0≤x≤2l

As B0,B≠0, we get C=0C=0

Using these values of A,A, p,p, CC in (6), the solution reduces to

y(x,t)=ksinnπ2lsinnπat2ly(x,t)=k\sin \frac{nπ}{2l}\sin \frac{nπat}{2l}

(7)----(7)

where n=0,1,2,3...n=0,1,2,3...\infty

The most general solution of Eq.(1) is

y(x,t)=n=1λnsinnπ2lcosnπat2ly(x,t)=\sum_{n=1}^{\infty}\lambda_{n}\sin \frac{nπ}{2l}\cos \frac{nπat}{2l} (8)----(8)

Differentiating both sides of (8) partially with respect to t, we have

δyδt(x,t)=\frac{\delta y}{\delta t} (x,t)= n=1(nπa2l.λn)sinnπx2lcosnπat2l\sum_{n=1}^{\infty}(\frac{nπa}{2l}.\lambda_{n})\sin \frac{nπx}{2l}\cos \frac{nπat}{2l} (9)----(9)

Using boundary condition (5) in (9), we have

n=1(nπa2lλn)sinnπx2l=k(2lxx2)\sum_{n=1}^{\infty}(\frac{nπa}{2l}\lambda_{n})\sin \frac{nπx}{2l}=k(2lx-x²)

forfor 0x2l0≤x≤2l

=n=1bnsinnπx2l=\sum_{n=1}^{\infty}b_{n}\sin \frac{nπx}{2l}

Which is Fourier half-range sine series of k(2lxx2)k(2lx-x²) in (0,2l).(0,2l).

Comparing like terms,we get

nπa2l.λn=bn=\frac{nπa}{2l}.\lambda_{n}=b_n=

22l02lf(x)sinnπxldx,\frac{2}{2l}\int_{0}^{2l}f(x)\sin \frac{nπx}{l}dx, by Euler's formula

=22l02lk(2lxx2)sinnπx2ldx=\frac{2}{2l}\int_{0}^{2l}k(2lx-x²)\sin \frac{nπx}{2l}dx

=\frac{2k}{n\pi a}[\,(2lx-x^{2})[\,\frac{-cos\frac{n\pi x}{2l}}{\frac{n\pi}{2l}}]\,-(2l-2x[\,\frac{-sin\frac{n\pi x}{2l}}{\frac{n^{2}\pi^{2}}{4l^{2}}}]\,\\+(-2)[\,\frac{cos\frac{n\pi x}{2l}}{\frac{n^{3}\pi^{3}}{8l^{3}}}]\,]\,_{0}^{2l}

=32kl3n4π4a[1(1)n]=\frac{32kl^{3}}{n^{4}\pi^{4}a}[1-(-1)^{n}]


={0,ifniseven32kl3n4π4a,ifnisodd=\begin{cases}0,\hspace{0.4cm}if\hspace{0.1cm} n \hspace{0.1cm}is\hspace{0.1cm} even\\\\\\ \frac{32kl^{3}}{n^{4}\pi^{4} a},\hspace{0.4cm}if\hspace{0.1cm} n \hspace{0.1cm}is\hspace{0.1cm} odd\end{cases}


Using this value of λn\lambda_n in (8), the required solution is

y(x,t)=y(x,t)= 64kl3π4an=11(2n1)4sin(2n1)πx2lcos(2n1)πat2l\frac{64kl³}{π⁴a}\sum_{n=1}^{\infty}\frac{1}{(2n-1)⁴}\sin \frac{(2n-1)πx}{2l}\cos \frac{(2n-1)πat}{2l}



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