Question

A string is stretched and fastened to two points at a distance 0

l

0 apart. Motion is started by

displacing the string in the form y = k sin πx

l

from which it is released at time t = 0. Show

that the displacement of any point on the string at a distance x from one end at time t is

given by k sin πx

l

cos πax

l

Answer 1

The equation of string is given by -

$\dfrac{\partial^{2}y}{\partial t^{2}}=a^{2}\dfrac{\partial^{2}y}{\partial x^{2}}$ $.................................1)$

Sine the string is stretched between the two points (0,0) and $(l,0)$ , hence the displacement of the string at these point will be zero .

$\therefore y(0,t)=0............................................2)$

$and \ y(l,t)=0...............................................3)$

since the string is released from rest hence its initial velocity will be zero .

$\therefore \dfrac{\partial y}{\partial t}=0\ at\ t=0 .....................................................4)$

since , the string is displaced from the initial position at time t=0 hence the initial displacement is given is

$y(x,0)=Ksin{\dfrac{\pi x}{l}}...........................................5)$

conditions (2), (3),(4),(5) are the boundary conditions

Let us now proceed to solve equation (1),

Let $y=XT...............................................6)$

where X is a function of $x\ only$ and T is a function of t only .

$\dfrac{\partial y}{\partial t}=\dfrac{\partial (XT)}{\partial t}=X\dfrac{dT}{dt}$

$\dfrac{\partial^{2} y}{\partial t^{2}}=\dfrac{\partial(X\dfrac{dT}{dt})}{\partial t}=X\dfrac{d^{2}T}{dt^{2}}$

similarly, $\dfrac{\partial^{2} y}{\partial t^{2}}=T\dfrac{d^{2}X}{dx^{2}}$

substituting the above in equation (1) , we get

$X\dfrac{d^{2}T}{dt^{2}}=a^{2}T\dfrac{d^{2}X}{dx^{2}}$ $\implies$ $XT^{''}=a^{2}TX^{''}$

Now making some cases , we get ,

$Case \ 1)\dfrac{1}{a^{2}}\dfrac{T^{''}}{T}=\dfrac{X^{''}}{X}=-P^{2}$

$i)$ $\dfrac{1}{a^{2}}\dfrac{T^{''}}{T}=-p^{2}$

$\dfrac{d^{2}T}{dt^{2}}+a^{2}p^{2}T=0$

Auxiliary equation is given by -

$m^{2}+c^{2}p^{2}=0$

$m=$ ${\pm}api$

$\therefore CF=c_1cosapt+c_2sincpt$

$PI=0$

$\therefore CF=c_1cos\ apt+c_2 sin \ {apt}$

$T=CF+PI$

$=c_1cos\ apt+c_2 sin \ {apt}.....................................7)$

$ii)$ $\dfrac{X''}{X}=-p^{2}$ $\implies$ $\dfrac{d^{2}X}{dx^{2}}+p^{2}X=0$

$\dfrac{X''}{X}=-p^{2}$

Auxiliary equation is -

$m^{2}+p^{2}=0$

$m={\pm}pi$

$\therefore C.F=c_{3}cos(px)+c_4sin(px)$ $=X..............................(8)$

hence , $y(x,t)=$ $(c_1cos\ apt+c_2 sin \ {apt})(c_{3}cos(px)+c_4sin(px))................................9)$

$Case \ 2)\dfrac{1}{a^{2}}\dfrac{T^{''}}{T}=\dfrac{X^{''}}{X}=p^{2}$

$i)$ $\dfrac{1}{a^{2}}\dfrac{T^{''}}{T}=p^{2}\implies\ \dfrac{d^{2}T}{dt^{2}}-a^{2}p^{2}T=0$

Auxiliary equation is - $m^{2}-p^{2}a^{2}=0\implies\ m={\pm}pa$

$\therefore$ $CF=c_5e^{apt}+c_6e^{-apt}$

$PI=0$

$\therefore$ $T=c_5e^{apt}+c_6e^{-apt}$

$ii)$ $\dfrac{X''}{X}=p^{2}$

$\dfrac{d^{2}X}{dx^{2}}-p^{2}X=0$

Auxiliary equation is -

$m^{2}-p^{2}=0\implies m={\pm}p$

$\therefore CF=c_{7}e^{px}+c_8e^{-px}$

$PI=0$

$\therefore$ $X=$ $c_{7}e^{px}+c_8e^{-px}$

hence , $y(x,t)=(c_5cose^{pat}+c_6e^{-cpt})(c_7e^{px}+c_8e^{-px})...................10)$

$Case\ III) \dfrac{1}{a^{2}}\dfrac{T^{''}}{T}=\dfrac{X^{''}}{X}=0(say)$

$(i)$ $\dfrac{1}{a^{2}}\dfrac{T^{''}}{T}=0\implies T^{''}=0$ or $\dfrac{d^{2}T}{dt^{2}}=0$

Auxiliary equation is

$m^{2}=0\implies m=0,0$

$\therefore C.F=c_9+c_{10}t$

$PI=0$

$\therefore$ $T=c_9+c_{10}t$

$ii)$ $\dfrac{X^{''}}{X}=0(say)\implies$ $\dfrac{d^{2}X}{dx^{2}}=0$

Auxiliary equation is -

$m^{2}=0\implies m=0,0$

$\therefore$ $CF=c_{11}+c_{12}x$

$PI=0$

$\therefore$ $X=$ $c_{11}+c_{12}x$

hence , $y(x,t)=(c_9+c_{10}x)(c_{11}+c_{12}x)........................................................11)$

Out of these three above solutions (9),(10)and (11), we have to choose the solution which is consistent with the physical nature of the problem . since we are dealing with problem on vibrations , the solution must contain periodic solution . hence the solutions which contains trignometric terms must be the solutions of given equations .

hence , solution (9) is the general solution of one -dimension wave equation given by the equation (1),

$now , y(x,t)=(c_1cos\ {apt}+c_{2}sin\ {apt})(c_3cos\ px+c_4sin \ px)$

Applying the boundary condition ,

$y(0,t)=0=$ $(c_1cos\ {apt}+c_{2}sin\ {apt})c_{3}$

$\implies$ $c_{3}=0$

$\therefore$ $from \ (9),$ $y(x,t)=(c_1cos\ {apt}+c_{2}sin\ {apt})c_{4}sin\ px$

again , now using the boundary conditions ,

$y(x,t)=0=(c_1cos\ {apt}+c_{2}sin\ {apt})c_{4}sin\ pl$

$\implies$ $sinpl=0=sin{n{\pi}}(n\isin I)$

$\therefore$ $p=\dfrac{n{\pi}}{l}$

hence from (12), $y(x,t)=(c_1cos\dfrac{n{\pi}at}{l}+c_2sin\dfrac{n{\pi}at}{l})$$c_{4}sin{\dfrac{n{\pi}x}{l}}$

now \ $(\dfrac{\partial y}{\partial t})_{t=0}=\dfrac{n{\pi}a}{l}[-c_{1}sin\dfrac{n{\pi}at}{l}+c_{2}sin\dfrac{n{\pi}at}{l}]c_4sin\dfrac{n{\pi}x}{l}$

now again using boundary conditions ,

At t=0 ,

$(\dfrac{\partial y}{\partial t})_{t=0}=0=\dfrac{n{\pi}c}{l}[c_2c_4sin\dfrac{n{\pi}x}{l}]$

$\implies$ $c_{2}=0$

$\therefore$ $from (13), \ y(x,t)=c_1c_4cos\dfrac{n{\pi}at}{l}sin\dfrac{n{\pi}x}{l}$

$y(x,0)=ksin\dfrac{{\pi}x}{l}=c_1c_4sin\dfrac{n{\pi}x}{l}$

where $c_{1}c_{4}=k, n=1$

hence from (14), $y(x,t)=kcos\dfrac{{\pi}ax}{l}sin\dfrac{{\pi}x}{l}$ which is required solution .