Fourier Series: Solutions of One Dimensional Wave Equation - Vibrating String with Non Zero Initial Velocity

ONE DIMENSIONAL WAVE EQUATION: VIBRATING STRING WITH NON-ZERO INITIAL VELOCITY

Problems on vibrating string with non‒zero initial velocity.

Type 2. Vibrating string with non‒zero initial velocity :

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

(i) y (0,t) = 0

(ii) y (l,t)  = 0

(iii) y (x, 0) = 0

(iv) ∂y/∂t _{(x, 0)} = f(x)

The suitable solution is

 y (x, t) = (c₁ cos px + c₂ sin px) (c₃ cos p at + c₄ sin p at)     ….. (1)

Apply condition (i), we get c₁ = 0

Apply condition (ii), we get p = nπ / l

Apply condition (iii), we get c₃ = 0

The most general solution is at

Apply condition (iv), we get

Substitute in (2) we get the general solution.

Example 15: A tightly stretched string with fixed end points x=0 and x = l is initially at rest in its equilibrium position. If it is set vibrating string giving each point a velocity λx (l‒x) show that the displacement is

Solution: The wave equation is 

From the given problem we get the following boundary and initial conditions,

(i) y (0, t) 0 for all t > 0

(ii) y (l, t) = 0 for all t > 0

(iii) y (x, 0) = 0, 0 <x< l

(iv) ( ∂y/∂t )_{(x, 0)} = λx (l − x), 0 < x < l

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

y (x, t) = (c₁ cospx + c₂sin px) (c₃ cos p at + c₄ sin p at)         ... (1)

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

 y (0, t) = c₁ (c₃ cos p at + c₄ sin pat) = 0

Here, c₃ cosp at + c₄ sin p at ≠ 0       [ It is defined for all t]

Therefore, we get

 c₁ = 0

Substitute c₁ = 0 in (1), we get

y (x, t) = c₂ sin px (c₃ cos p at + c₄ sin p at)          ... (2)

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

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

Here, (c₃ cos p at + c₄ sin p at) ≠ 0      [. It is defined for all t]

Therefore, either c₂ = 0 or sin pl = 0

Suppose, we take c₂ = 0 and already we have c₁ = 0 then we get a trivial solution.

Therefore, we consider c₂≠ 0 and sinpl = 0

 sin pl = 0

 pl = nπ

     ['. sin nπ =0]

P = nπ / l       [n being an integer]

Now, substituting p = nπ / l in equation (2), we get

Applying condition (iii) in equation (3), we get

 y (x, 0) = c₂ sinn(xnπ/l) c₃ = 0

 c₂ c₃ sin (xnπ/l) = 0

Here, sin (xnπ/l) ≠ 0   ['. It is defined for all x]

 c₂ ≠ 0     ['. If c₂ = 0 we already explained]

Therefore, c₃ = 0

Substitute c₃ = 0 in equation (3), we get

To find B_n: Expand λx(1‒x) in a half range Fourier sine series in the interval (0, l)

From (6) and (8), we get  b_n = B_n

Example 16: If a string of length a is initially at rest in its equilibrium position and each of its points is given a velocity kx (a‒x), determine the displacement function.

Solution: In Example 15 Put l=a and λ = k

Example 17: A tightly stretched string with fixed end points x = 0 and x = L, is initially in its equilibrium position. If it is set vibrating giving each point a velocity 3x (L x), find the displacement.

Solution: In Example 15

Here, Put l = L and λ=3

Example 18: A tightly stretched string with fixed end points x=0, and x = L is initially in a position given by y = Lx‒x² it is released from rest from this position, find the displacement y(x, t)

Solution: In Example 15

Here, Put l = L and λ=1

Example 19: If a string of length l is initially at rest in its equilibrium position and each of its points is given the velocity v₀ sin³ πx/l, 0 < x <l, determine the displacement of a point distant x from one end at time 't'.

Solution: The wave equation is 

From the problem we get the following boundary and initial conditions

(i) y (0, t) = 0, for all t > 0

(ii) y (l, t) = 0, for all t > 0

(iii) y (x, 0) = 0, 0 < x < 1

(iv) ( ∂y/∂t )_{(x, 0)} = v₀ sin³ πx/l , 0 < x < l

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

y (x, t) = (c₁ cos px + c₂ sin px) (c₃ cos p at + c₄ sin p at)         …….. (1)

Equations from (1) to (6) is same as Example 15

Example 20: A string of length l is initially at rest in its equilibrium position and motion is started by giving each of its points a velocity given by v 

Find the displacement function y (x, t).

Solution:

The wave equation is 

From the given problem we get the following boundary and initial conditions,

(i) y (0, t) for all t ≥ 0

(ii) y (l, t) = 0, for all t ≥ 0

(iii) y (x, 0) = 0

(iv) ( ∂y/∂t )_{(x, 0)} 

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

 y(x, t) = (c₁ cos px + c₂ sin px) (c₃ cos p at + c₄ sin p at)    ………… (1)

Equations from (1) to (5) is same as Example no. 15

To find B_n: Expand f(x) in a half range Fourier sine series in the interval (0, 1)

Example 21: If the string of length l is initially at rest in equilibrium position and each of its points is given the velocity  where 0 < x < 1 at t = 0, determine the displacement function y(x, t)

Solution: The wave equation is 

From the given problem, we get the following boundary and initial conditions,

(i) y (0, t) = 0 for all t≥0

(ii) y (l, t) = 0 for all t≥0

(iii) y (x, 0) = 0, for all t ≥ 0

(iv) ( ∂y/∂t )_{(x, 0)} = , 0 < x < l

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

 y (x, t) = (c₁ cos px + c₂ sin px) (c₃ cos p at + c₄ sin p at)             ………..(1)

Equations from (1) to (5) is same as Example no. 15

Example 22: A string is stretched between two fixed points at a distance 2l apart and the points of the string are given initial velocities v where

 v = cx/l in 0 < x <l

= c/l (2l‒x) in l<x<2l

 x being the distance from one end point. Find the displacement of the string at any subsequent time.

Solution: The wave equation is 

From the given problem, we get the following boundary and initial conditions,

(i) y (0, t) = 0 for all t≥ 0

(ii) y (2l, t) = 0 for all t≥0

(iii) y (x, 0) = 0

(iv) ( ∂y/∂t )_{(x, 0)}

Equations from (1) to (5) is same as Example 15

To find B_n expand f(x) in a Half range Fourier sine series in the interval (0, 21)

EXERCISE

1. A taut string of length 20 cm fastened at both ends is displaced from its position of equilibrium by imparting to each of its points an initial velocity given by

 x being the distance from one end. Determine the displacement at any subsequent time.

2. A string is stretched between two fixed points at a distance of 60 cm and the points of the string are given initial velocities v, where

u = λx/30, in 0 < x < 30

= λ/30 (60‒x), in 30 < x < 60

 x being the distance from an end point. Find the displacement of the string at any time.

3. Solve the boundary value problem 

 y (0,t) = y (5, t) =  0,

 y (x, 0) =0,

( ∂y/∂t )_t=0 = f(x)

If (i) f(x) = 5 sin πx (i) f(x) = 3 sin 2πx ‒ 2 sin 5 πx

4. Solve ∂²y/∂t² = 4 ∂²y/∂x² , y(0,t) = y(5,t) = 0, y (x, 0) = 0,

 ∂y/∂t (x, 0) = 3 sin 2πx ‒ 2 sin 5πx

ANSWERS