Solutions Of One Dimensional Wave Equation

SOLUTIONS OF ONE DIMENSIONAL WAVE EQUATION

Equation of a vibrating string :- One dimensional wave equation: Consider an elastic string tightly stretched between two points O and A. Let O be the origin and OA as x-axis. On giving a small displacement to the string, perpendicular to its length (parallel to the y-axis). Let y be the displacement at the point P (x, y) at any time. The wave

Example 3.3.1: Obtain the solution of one dimensional wave equation:

p

Out of the three mathematically possible solutions derived, we have to choose the solution which is consistent with the physical nature of the problem and the given boundary conditions. In the case of vibration of an elastic string, y (x, t) representing the displacement of the string at any point x, y must be periodic in t. Hence solution (9), which consists of periodic functions in t is the suitable solution of the problems on vibration of strings. The arbitrary constants in the suitable solution are found out by using the boundary conditions of the problem. In problems, we directly assume that (9) is the suitable solution of vibration of string problems.

The suitable solution (9) which is periodic in 't is periodic in 'x' also.

Example 3.3.2: Derive D' Alembert's solution of the wave equation:

This solution is called D' Alemberts solution of the one dimensional wave equation.

Example 3.3.3: State the assumptions made in the derivation of one dimensional wave equation.

Solution: The wave equation 

This gives y (x, t), the transverse vibration of a string stretched to a constant tension T. In deriving this equation, we make the following assumptions.

1. The motion takes place entirely in one plane i.e., xy plane.

2. We consider only transverse vibrations, the horizontal displacement of the particles of the string is negligible.

3. The tension T is constant at all times and at all points of the deflected string.

4. T is considered to be so large compared with the weight of the string and hence the force of gravity is negligible. 租

5. The effect of friction is negligible.

6. The string is perfectly flexible, i.e., it can transmit tension but not bus bending or shearing forces.

7. The slope of the deflection curve at all points and at all instants is so small that sin a can be replaced by a, where a is the inclination of the tangent to the deflection curve.

Example 3.3.4: Derive one dimensional wave equation :

Solution : Consider a tightly stretched elastic string of length with its end points fixed. Let the string be released from rest and allowed to vibrate. The problem is to determine the deflectior y (x, t) at any point x and at any time t > 0.

For deriving the p.d.e. we make the following assumption.

See Example 3.3.3.

To obtain the differential equation

Take one end of the string as origin O and take X-axis along the string.

We assume that the motion takes place entirely in the XY plane. Consider the forces acting on a small portion PQ of the string where P is (x, y) and Q is (x + ∆x, y + ∆y)

By assumption, the string does not offer resistance to bending and shearing hence the tension at each point of the string is tangential to the curve of the string.

Jon Let T₁ and T₂ be the tensions at the points P and Q.

Letφ and φ + ∆φ be the angles made by the tangents at P and Q respectively with the X axis.

Let m be the mass per unit length of the string

Since, there is no motion in the horizontal portion of the string ast od lo noitsailoni the tension must be constant.

.. T1 cos φ = T₂ cos (φ+ ∆φ) which is a constant.

Sinceφ is small, cosy φand cos (φ+∆φ) are approximately equal to 1.

Thus, T₁ = T₂ = T, a constant.

The vertical component of the force acting on the element PQ is

T sin (φ+∆φ) - T sinφ= T (φ + ∆ φ) − Tφ

 ['. sin φ  = φwhereφ is small]

The acceleration of the element in the y direction is 

Hence, by Newton's second law of motion,

 = T ∆φ where m is the mass per unit length of the string.

as ∆S is the length of the element PQ.

Taking limit as as ∆S→ 0, we get

This partial differential equation is known as the one dimensional wave equation. It is a homogeneous of second order.

Problem on vibrating string with zero initial velocity.

Type 1. Vibrating string with zero initial velocity:

The boundary and initial conditions of the deflection y (x, t) are

The suitable solution is

 (x,t) = (c₁ cos px + c₂ sin px) (c₃ cos p at + c₄ sin p at)

Apply condition (i), we get c₁ = 0

Apply condition (ii), we get

Apply condition (iii), we get c₄ = 0

The most general solution is

Example 3.3.6 : A string is stretched and fastened to two points x = 0 and x = 1 apart. Motion is started by displacing the string into the form y = k (lx - x) from which it is released at time t = 0. Find the displacement of any point on the string at a distance of x from one end at time t. 

Solution: The wave equation is        

From the given problem, we get the following boundary & initial conditions.

Now, the suitable solution which satisfies our boundary conditions is given by

Applying condition (i) in equation (1), we get

y (0,t) = (c₁+0) (c₃ cos p at + c₄ sin p at)……………………....(1)

Substitute, c₁ = 0 in equation (1), we get (vi) noilib

y (x, t) C₂ sin px (c₃ cos p at + c₄ sin p at)

Applying condition (ii) in equation (2), we get

y (l, t) = c₂ sin pl (c₃ cos p at + c₄ sin pat) = 0

Here, [c₃ cos pat + c₄ sin pat] 0 [. it is defined for all t]

Therefore, either c2 bart 0 =0

Example 3.3.7: Find the displacement of a string of length '7' vibrating between fixed end points with initial velocity zero and initial displacement is given by

Example 3.3.8 : A string of length 21 is fastened at both ends. The mid point of the string is taken to a height b and then released from rest in that position. Show that the displacement is

Example 3.3.9 : A tightly stretched string of length has its ends fastened at x = 0 and x = 1. The mid point of the string is then taken to a height hand then released from rest in that position. Obtain an expression for the displacement of the string at any subsequent time.

Example 3.3.10 : The points of trisection of a string of length / with fixed ends aside through a distance h on opposite sides of the position of equilibrium and the string is released from rest. Find an expression for the displacement of the string at any subsequent time. Also show that the mid point of the string always remains at rest.

Solution: Taking the end points as origin O and A on the x axis, the initial position of the string is given in the figure.

Example 3.3.11: If a string of length /is released from rest in the position ppppppppppppp show  that the motion is described by the equation.

Example 3.3.12: A tightly stretched string with fixed end points x=0 and x = / initially displaced in a sinusoidal arc of length yo and then released from rest. Find the displacement y at any distance x from one end at time t.

Example 3.3.13 : An elastic string is stretched between two points at a distance л apart. In its equilibrium position the string is in the shape of the curve f (x) = k (sin x – sin³ x). Obtain y(x, t) the vertical 

Example 3.3.14: A string of length l has its ends x = 0, x = l fixed. The point where x = l/3 is drawn aside a small distance h, the displacement y (x, t) satisfies Find y(x, t) at any time t.