Answer to Question #171473 in Differential Equations for ganesh

Question

A tightly stretched string of length ‘l’ has its ends fastened at x=0, x=l. The midpoint

of the string is then taken to height ‘h’ and then released from rest in that

position. Find the initial displacement y(x,0) of the string.

Answer 1

When string is released, then it will vibrate. Let the equation which represents the displacement is given by, $y(x,t)= Asin(\omega t +kx+\phi)$

where omega is angular velocity, k is wave number, phi is initial phase.

According to the question, at end points x=0 and x=l, displacement is zero and maximum displacement is at mid point that is h.

So, $A=h$

Now, since string is held at end points so it represents half wavelength of the wave. Then $k=\frac{2\pi}{\lambda}, \lambda =l/2 \implies k=\frac{4\pi}{l}$

Since we are asked at t=0, so equation will look like $y(x,0)= hsin(kx+\phi)$

For phase, at x=0 and x=l, displacement is zero, so

$0= hsin(\phi)$ and. $0 = hsin(\frac{4π}{l}l+\phi)= hsin(4π+\phi)$

Solving these, we get $\phi = 0$

Hence final equation will be, $y(x,0)= hsin( \frac{4π}{l} x)$