Question

3. A tightly stretched string of length / with fixed ends is initially in equilibrium position. It is set vibrating by giving each point a velocity v 0 sin^ 3 pi x l . Find the displacement y(x, t) .​

Answer 1

Answer:

Solution of the wave equation

The wave equation is

Let y = X(x) . T(t) be the solution of (1), where „X‟ is a function of „x‟ only and „T‟ is a function of „t‟ only.

Of these three solutions, we have to select that particular solution which suits the physical nature of the problem and the given boundary conditions. Since we are dealing with problems on vibrations of strings, „y‟ must be a periodic function of „x‟ and „t‟.

Now the left side of (2) is a function of „x‟ only and the right side is a function of „t‟ only.

Since „x‟ and „t‟ are independent variables, (2) can hold good only if each side is equal to a constant.

Hence the solution must involve trigonometric terms.

Therefore, the solution given by (5),

i.e, y = (c5 coslx + c6 sin lx) (c7 cosalt+ c8 sin alt)

is the only suitable solution of the wave equation.

llustrative Examples.

Example 1

If a string of length ℓ is initially at rest in equilibrium position and each of its points is given the velocity

Solution

The displacement y(x,t) is given by the equation

The boundary conditions are

i. y(0,t) = 0, for t ³0.

ii. y(ℓ,t) ³0.= 0, for t

y(x,0) = 0, for 0 £x £ℓ.

Since the vibration of a string is periodic, therefore, the solution of (1) is of the form

y(x,t) = (Acoslx + Bsinlx)(Ccoslat + Dsinlat) ------------(2)

Using (i) in (2) , we get

0 = A(Ccoslat + Dsinlat) , for all t ³0.

Therefore, A = 0

Hence equation (2) becomes

y(x,t) = B sinlx(Ccoslat + Dsinlat) ------------ (3)

Using (ii) in (3), we get

0 = Bsinlℓ (Ccoslat+Dsinlat), for all t ³0, which gives lℓ = np

Hence, l= np / l , n being an integer.

Using condition (iv) in the above equation, we get

Example 2

A tightly stretched string with fixed end points x = 0 & x = ℓ is initially at rest in its equilibrium position . If it is set vibrating by giving to each of its points a velocity

¶y/¶t = kx(ℓ-x) at t = 0. Find the displacement y(x,t).

Solution

The displacement y(x,t) is given by the equation

The boundary conditions are

Example 3

A tightly stretched string with fixed end points x = 0 & x = ℓ is initially in a position given by y(x,0) = y0sin3(px/ℓ). If it is released from this position, find the displacement y at any time and at any distance from the end x = 0 .

Solution

The displacement y(x,t) is given by the equation

The boundary conditions are

(i) y(0,t) = 0, "t ³0.

(ii) y("tℓ³,t)0. = 0,

(iv) y(x,0) = y0 sin3((px/ℓ), for 0 < x < ℓ.

The suitable solution of (1) is given by

y(x,t) = (Acoslx + Bsinlx)(Ccoslat + Dsinlat) ------------(2)

Using (i) and (ii) in (2) , we get

Example 4

A string is stretched & fastened to two points x = 0 and x = ℓ apart. Motion is started by displacing the string into the form y(x,0) = k(ℓx-x2) from which it is released at time t = 0. Find the displacement y(x,t).

Solution

The displacement y(x,t) is given by the equation

The boundary conditions are

(i) y(0,t) = 0, "t ³0.

(ii) y(ℓ,t) "t ³=0. 0,

The suitable solution of (1) is given by

y(x,t) = (Acoslx + Bsinlx)(Ccoslat + Dsinlat) ------------(2)

Using (i) and (ii) in (2) , we get

Example 5

A uniform elastic string of length 2ℓ is fastened at both ends. The midpoint of the string is taken to the height „b‟ and then released from rest in that position . Find the displacement of the string.

Solution

The displacement y(x,t) is given by the equation

The suitable solution of (1) is given by

y(x,t) = (Acoslx + Bsinlx)(Ccoslat + Dsinlat) ------------(2)

The boundary conditions are

(i) y(0,t) = 0, "t ³0.

(ii) y(ℓ,t) "t ³=0. 0,

[Since, equation of OA is(y- b)/(oy-b)== (x(b/-ℓ)/(2ℓ-ℓ)x)]ℓ

Using conditions (i) and (ii) in (2), we get

Here B can not be zero, therefore D = 0.

Hence equation (3) becomes

Example 6

A tightly stretched string with fixed end points x = 0 & x = ℓ is initially in the position y(x,0) = f(x). It is set vibrating by giving to each of its points a velocity ¶y/¶t = g(x) at t = 0 . Find the displacement y(x,t) in the form of Fourier series.

Solution

The displacement y(x,t) is given by the equation

The boundary conditions are

(i) y(0,t) = 0, "t ³0.

(ii) y(ℓ,t) "t ³=0. 0,

(iii) y(x,0) = f(x) , for 0 £x £ℓ.

=g(x) , for 0 £x £ℓ. ¶t t

The solution of equation .(1) is given by

y(x,t) = (Acoslx + Bsinlx)(Ccoslat + Dsinlat) ------------(2)

where A, B, C, D are constants.

Applying conditions (i) and (ii) in (2),

Answer 2

Concept Introduction:-

It could take the shape of a word or a numerical representation of a quantity's arithmetic value.

Given Information:-

We have been given that a tightly stretched string of length / with fixed ends is initially in equilibrium position. It is set vibrating by giving each point a velocity v $0$ $sin^ 3 \pi x l$

To Find:-

We have to find that the displacement $y(x, t)$.​

Solution:-

According to the problem

The displacement $\mathrm{y}(\mathrm{x}, \mathrm{t})$ is given by the equation

$\frac{\partial^{2} y}{\partial t^{2}}=a^{2} \frac{\partial^{2} y}{\partial x^{2}}$

The boundary conditions are

$(i)$  $y(0, t)=0$, for $t \geq 0$

$(ii)$ $y(\ell, t) \geq 0 .=0$, for $t$ $y(x, 0)=0$, for $0 \leq x \leq \ell$

$iv$. $\left(\frac{\partial y}{\partial t}\right)_{t=0}=v_{0} \sin \frac{\pi x}{\ell}$, for $0 \leq x \leq \ell$.

Since the vibration of a string is periodic, therefore, the solution of $(1)$ is of the form $y(x, t)=(A \cos \lambda x+B \sin \lambda x)(C \cos \lambda a t+D \sin \lambda a t) \cdots(2)$

Using $(i)$ in$(2)$, we get

$0=\mathrm{A}(\mathrm{C} \cos \lambda \mathrm{at}+\mathrm{D} \sin \lambda \mathrm{at}),$ for all $\mathrm{t} \geq 0$

Therefore, $\quad \mathrm{A}=0$

Hence equation $(2)$ becomes

$y(x, t)=B \sin \lambda x(C \cos \lambda a t+D \sin \lambda a t) \cdots(3)$

Using $(ii)$ in $(3)$, we get

$0=$ Bsin $\lambda \ell \quad(C \cos \lambda$ at $+$ Dsin $\lambda$ at), for all $t \geq 0$, which gives $\lambda \ell=n \pi$

Hence, $\quad \lambda=\mathrm{n} \pi / 1, \mathrm{n}$ being an integer.

Thus, $y(x, t)=B \sin \frac{n \pi x}{\ell} \quad\left(C \cos \frac{n \pi a t}{\ell}+D \sin \frac{n \pi a t}{\ell}\right) \ldots(4)$

Using $(iii)$ in $(4)$, we get

$0=B \sin \frac{n \pi x}{\ell} C$

which implies $\mathrm{C}=0$

$\begin{aligned}\therefore \quad y(x, t) &=B \sin \frac{n \pi x}{\ell} \cdot D \sin \frac{n \pi a t}{\ell} \\&=B_{1} \sin \frac{n \pi x}{\ell} \cdot \sin \frac{n \pi a t}{\ell}, \text { where } B_{1}=B D .\end{aligned}$

The most general solution is

$y(x, t)=\sum_{n=1}^{\infty} B_{n} \sin \frac{n \pi x}{\ell} \sin \frac{n \pi a t}{\ell} \text {.......5 }$

Differentiating $(5)$ partially w.r.t t, we get

$\frac{\partial \mathrm{y}}{\partial \mathrm{t}}=\sum_{\mathrm{n}=1}^{\infty} \mathrm{B}_{\mathrm{n}} \sin \frac{\mathrm{n} \pi \mathrm{x}}{\ell} \cdot \cos \frac{\mathrm{n} \pi \mathrm{at}}{\ell} \cdot \frac{\mathrm{n} \pi \mathrm{a}}{\ell}$

Using condition $(iv)$ in the above equation, we get

$v_{0} \sin \frac{\pi x}{\ell}=\sum_{n=1}^{\infty} \quad B_{n} \cdot \frac{n \pi a}{\ell} \cdot \sin \frac{n \pi x}{\ell}$

Equating like coefficients on both sides, we get

$\mathrm{B}_{1} \frac{\pi \mathrm{a}}{\ell}=\mathrm{v}_{\mathrm{o}}, \mathrm{B}_{2} \cdot \frac{2 \pi \mathrm{a}}{\ell}=0, \mathrm{~B}_{3} \frac{3 \pi \mathrm{a}}{\ell}=0, \ldots .$

i.e, $\mathrm{B}_{1}=\frac{\mathrm{v}_{0} \mathrm{l}}{\pi \mathrm{a}}, \quad \mathrm{B}_{2}=\mathrm{B}_{3}=\mathrm{B}_{4}=\mathrm{B}_{5}=\cdots \cdots \cdot 0$ .

Substituting these values in $(5)$, we get the required solution.

i.e, $y(x, t)=\frac{v_{0} \ell}{\pi a} \quad \sin \frac{\pi x}{\ell} \quad \sin \frac{\pi a t}{\ell}$

Final Answer:-

The correct answer is $y(x, t)=\frac{v_{0} \ell}{\pi a} \quad \sin \frac{\pi x}{\ell} \quad \sin \frac{\pi a t}{\ell}$.

#SPJ2