Question

A tightly stretched string with fixed end points x=0 and x=1 is initialy in a position given by y=y0 sin^3(pi.x/l). it is released from rest from the initial position. find the displacement y(x,t)

Answer 1

Answer on Question #74059 – Math – Differential Equations

Question

A tightly stretched string with fixed end points $x=0$ and $x=1$ is initially in a position given by $y = y_0 \sin^3 \left( \frac{\pi x}{l} \right)$; it is released from rest from the initial position. Find the displacement $y(x,t)$.

Solution

The equation of the string is $\frac{\partial^2 y}{\partial t^2} = c^2 \frac{\partial^2 y}{\partial x^2}$ (1)

Let $y = X(x)T(t)$

$X$ is a function of $x$ and $T$ is the function of $t$.

$$\frac{\partial y}{\partial t} = \frac{\partial (X T)}{\partial t} = X T' \quad \frac{\partial^2 y}{\partial t^2} = X T'$$

Similarly we calculate $\frac{\partial^2 y}{\partial x^2}$ and we get

$$\frac{\partial y}{\partial x} = \frac{\partial (X T)}{\partial x} = X' T \quad \frac{\partial^2 y}{\partial x^2} = X'' T$$

Putting into equation (1), we get

$$X T'' = c^2 T X'$$ $$\frac{1}{c^2} \frac{T''}{T} = \frac{X''}{X} = K = (-p^2) \quad (say)$$

Using by separation of variable we solve these two cases separately

$$\frac{1}{c^2} \frac{T''}{T} = -p^2 \quad \text{and} \quad \frac{X''}{X} = -p^2$$ $$T'' = -c^2 p^2 T \quad \text{and} \quad X'' = -p^2 X$$ $$D' = \pm p c i \quad \text{and} \quad D = \pm p i$$ $$T = C_1 \cos c p t + C_2 \sin c p t \quad X = C_3 \cos p x + C_4 \sin p x$$

General solution of equation (3) and put into equation (1)

$$y(x,t) = (C_1 \cos c p t + C_2 \sin c p t) (C_3 \cos p x + C_4 \sin p x)$$

Given boundary conditions are

$$y (0, t) = 0, \quad y (l, t) = 0 \left(\frac {\partial y}{\partial t}\right) _ {t = 0} = 0, \quad y (x, 0) = y _ {0} \sin^ {3} \left(\frac {\pi x}{l}\right)$$

Applying boundaries conditions in equation (3*)

$$y (0, t) = 0 = \left(C _ {1} \cos c p t + C _ {2} \sin c p t\right) \left(C _ {3}\right)$$

So $C_3 = 0$

Equation (3*) will become $y(x,t) = (C_1\cos cpt + C_2\sin cpt)(C_4\sin px)$ (5)

Again $y(l,t) = 0$ $y(x,t) = (C_1\cos cpt + C_2\sin cpt)(C_4\sin pl)$

$\sin pl = 0 = \sin p\pi$

We get $p = \left(\frac{\pi x}{l}\right)$ $n \in I$

Hence from (5) $y(x,t) = (C_1\cos \frac{n\pi ct}{l} + C_2\sin \frac{n\pi ct}{l})(C_4\sin \frac{n\pi x}{l})$

$$\left(\frac {\partial y}{\partial t}\right) = \frac {n \pi c}{l} \left[ \left(- C _ {1} \sin \frac {n \pi c t}{l} + C _ {2} \cos \frac {n \pi c t}{l}\right) \left(C _ {4} \sin \frac {n \pi x}{l}\right) \right]$$

At $t = 0$

$$\left(\frac {\partial y}{\partial t}\right) = 0 = \frac {n \pi c}{l} \left[ C _ {2} C _ {4} \sin \frac {n \pi x}{l}) \right] \quad \text{hence} \quad c _ {2} = 0$$

Hence

$$y (x, t) = \left[ \left(C _ {1} C _ {4} \cos \frac {n \pi c t}{l} \sin \frac {n \pi x}{l}\right) \right], \quad y (x, t) = \left[ \left(b _ {n} \cos \frac {n \pi c t}{l} \sin \frac {n \pi x}{l}\right) \right] \quad \text{here} \quad b _ {n} = C _ {1} C _ {2}$$

Most general solution is

$$y (x, t) = \left[ \sum_ {n = 1} ^ {\infty} b _ {n} \cos \frac {n \pi c t}{l} \sin \frac {n \pi x}{l}) \right]$$ $$y (x, 0) = y _ {0} \sin^ {3} \left(\frac {\pi x}{l}\right) = \left[ \sum_ {n = 1} ^ {\infty} b _ {n} \sin \frac {n \pi x}{l} \right]$$ $$y _ {0} = \frac {1}{4} \left(3 \sin \frac {\pi x}{l} - \sin \frac {3 \pi x}{l}\right) = b _ {1} \sin \frac {1 \pi x}{l} + b _ {2} \sin \frac {2 \pi x}{l} + b _ {3} \sin \frac {3 \pi x}{l} + \dots$$

Comparing and we get

$$b _ {1} = \frac {3 y _ {0}}{4}, b _ {2} = 0, b _ {3} = \frac {- y _ {0}}{4}, b _ {4} = 0, b _ {5} = 0 \dots \dots .$$

Finally equation (6) will be

$$y (x, t) = \left[ \frac {3 y _ {0}}{4} \cos \frac {\pi c t}{l} \sin \frac {n \pi x}{l} - \frac {y _ {0}}{4} \cos \frac {3 \pi c t}{l} \sin \frac {3 \pi x}{l} \right]$$

which is the required displacement.

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