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

ONE DIMENSIONAL WAVE EQUATION: VIBRATING STRING WITH ZERO INITIAL VELOCITY

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

(i) y (0, t) = 0 [boundary condition]

(ii) y (l, t) = 0 [boundary condition]

(iii) ∂y/∂t (x, 0) = 0 [initial condition]

(iv) y (x, 0) = f(x) [initial condition]

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 c₄ = 0

The most general solution is

Apply condition (iv), we get

Substitute, C_n in (2) we get the general solution.

 

Example 6: A string is stretched and fastened to two points x = 0 and x= l apart. Motion is started by displacing the string into the form y= k(lx ‒ x²) 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.

(i) y (0, t) = 0 for all t > 0 [B.C]

(ii) y (l,t) = 0 for all t> 0 [B.C]

(iii) (∂y/∂t)_(x,0) = 0, 0 < x < l    [I.C]

(iv) y(x, 0) = k (lx = x²),  0 < x < l    [I.C]

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

y (x, t) = (c₁cos px + c₂sin px) (c₃ cos p at + c₄ sin p a t)     ... (1)

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

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

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

Therefore, we get c₁ = 0

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

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

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

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

Here, [c₃ cos p at + c₄ sin pa t] ≠ 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

 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

Before applying condition (iii), diff. (3) p.w.r.to 't', we get

Now applying condition (iii), we get

The most general solution is

Applying the condition (iv) in equation (5), we get

To find C_n: expand k(lx − x²) in a half‒range Fourier Sine Series in the interval (0, l)

 

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

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/∂t)_{(x, 0)} = 0, 0 ≤ x ≤ l

(iv) 

Equations from (1) to (5) is same as example no. 3

 (5) ⇒ y (x, t) = 

Now, apply condition (iv) in (5), we get

To find C_n: Expand the given function in a half range Fourier sine series in the interval (0,l)

 

Example 8: A string of length 2l 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

Solution:

The wave equation is ∂²y/∂t² = a² ∂²y/∂x²

 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

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)

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

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

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

Therefore, c₁ = 0

Substitute c₁ = 0 in equation (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(2l, t) = c₂ sin 2lp (c₃ cos p at + c₄ sin pat) = 0

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

Therefore, either c₂ = 0 or sin 2 pl = 0

Suppose, we take c₂ = 0 and already we have c₁ = 0

then we get a trivial solution,

Therefore,

c₂ ≠ 0

The only possibility is sin 2 pl = 0

i.c., 2pl = n π     

        [ sinn π = 0]

P = nπ / 2l

Substitute, P = nπ / 2l in equation (2), we get

To find C_n expand the given value in a half‒range Fourier sine series in the interval (0, L)

Here, L = 2l

Substitute the value of c_n in equation (5), we get

Note: sin (2n‒1) π/2 = (−1)ⁿ⁻¹

 

Example 9: A tightly stretched string of length l has its ends fastened at x = 0 and x = l. The mid point of the string is then taken to a height h and then released from rest in that position. Obtain an expression for the displacement of the string at any subsequent time.

Solution: The equation to be solved is ∂²y/∂t² = a² ∂²y/∂x²

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

 

Example 10: The points of trisection of a string of length l 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 time. Also show that t 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.

The displacement function y(x, t) of the string is the solution of wave equation.

∂²y/∂t² = a² ∂²y/∂x²

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/∂t)_{(x, 0)} = 0 for 0 ≤ x ≤ l

(iv) y (x, 0) = f(x) where f(x) is given in (A)

Equations from (1) to (5) is same as Example no. 6

Applying the boundary condition (iv) in (5), we get

To find C_n expand f (x) in a half range Fourier sine series in the intervel (0, l)

The displacement at the mid point is got by substituting

x = l/2 in (8) 

when x = l/2 sin (2nπx / l) = sin nπ = 0 for all integral values of n

y (l/2 , t) = 0.

Hence, there is no displacement at x = l/2

for all values of t.

  The mid point of string is at rest.

 

Example 11: If a string of length l is released from rest in the position y = 4λx(1−x) / l². Show that the motion is described by the equation.

Solution:

The wave equation is

 ∂²y/∂t² = a² ∂²y/∂x²

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/∂t)_{(x, 0)} = 0, 0 ≤ x ≤ l

(iv) y (x, 0) = f(x) where f(x) = 4λx(1−x) / l², 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. 6

Applying condition (iv) in equation (5), we getsim

To find C_n: Expand 4λx(1−x) / l² in a half range Fourier sine series in the interval (0, 1)

Substitute the value of C_n in (5), we get

 

Example 12: A tightly stretched string with fixed end points x=0 and x = l initially displaced in a sinusoidal arc of length y₀ and then released from rest. Find the displacement y at any distance x from one end at time t.

Solution: The wave equation is

∂²y/∂t² = a² ∂²y/∂x²

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/∂t)_{(x, 0)} = 0, 0 ≤ x ≤ l

(iv) y (x, 0) = f(x) where f(x) =  y₀ sin(πx/l), 0 ≤ x ≤ l

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

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

Equations from (1) to (5) is same as Example no. 6

 

Example 13: An elastic string is stretched between two points at a distance π apart. In its equilibrium position the string is in the shape of the curve f(x) = k (sin x − sin³x). Obtain y(x, t) the vertical displacement if y satisfies the equation ∂²y/∂t² = ∂²y/∂x².

Solution: The wave equation is

 ∂²y/∂t² = ∂²y/∂x²

The boundary & initial conditions are,

(i) y (0,t) = 0

(ii) y (π,t) = 0

(iii) (∂y/∂t)_{(x, 0)} = 0, 0 ≤ x ≤ π

(iv) y (x, 0) = k(sin x − sin³x), 0 ≤ x ≤ π

 = k[ sin x – 1/4(3sinx‒ sin3x)]

       sin3θ = 3sinθ ‒ 4sin³θ

       sin³θ = ¼ [3sinθ ‒ sin3θ]

= k/4 [ 4sinx ‒ 3sinx + sin3x ]

= k/4 [ sinx + sin3x ]

= k/4 sinx + k/4 sin3x

Equations from (1) to (5) is same as Example no. 6

(5) = y (x, t) ⇒ ^∞Σ_n=1 C_n sin nx cos nt       ………. (5)

Applying condition (iv) in (5), we get

y (x, 0) = ^∞Σ_n=1 C_n sin nx = k/4 sinx + k/4 sin3x

i.e.,

C₁ sin x + C₂ sin 2x + C₃ sin 3x + C₄ sin 4x + ….

= k/4 sinx + k/4 sin3x        ………(6)

Equating like coefficients on both sides of (6), we get

C₁ = k/4

C₂ = 0,

C₃ = k/4

C₄ = C₅ = 0  ...

Substituting these values in (5), we get

 y (x, t) = [ k/4 sin x cost ] + [ k/4 sin 3x cos 3t ]

 

Example 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 ∂²y/∂t² = a² ∂²y/∂x². Find y(x, t) at any time t.

Solution: The wave equation is ∂²y/∂t² = a² ∂²y/∂x²

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/∂t)_{(x, 0)} = 0 for 0 < x < l

Equation of OB is

Applying the boundary condition (iv) in equation (5), we get

To find C_n expand f(x) in a half‒range Fourier sine series in the interval (0, l)

 

EXERCISE

 

1. A uniform elastic string of length 60 cm is subjected to a constant tension of 2 kg. If the ends are fixed and the initial displacement is y(x, 0) = 60x‒x² for 0 < x < 60 while the initial velocity is zero, find the displacement function y (x, t).

2. A tightly stretched string with fixed end points x = 0 and x = l is initially in a position given by y (x, 0) = k[sin(πx/l) ‒ sin(2πx/l)]. If it is released from rest from this position find the displacement y at any distance x from one end at any time t.

3. A taut string of length 2l is fastened at both ends. The mid point of the string is taken to a height h and then released from rest in that position. Find the displacement of the string.

4. The points of trisection of a tightly stretched string of length 30 cm with fixed ends pulled aside through a distance of 1 cm 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 points of the string remains always at rest.

5. A string of length l is fastened at both ends. One end is taken as the origin and at a distance b from this end, the string is displaced a distance d transversely and is released from rest when it is in this position. Find the equation of the sub‒sequent motion.

6. A tightly stretched flexible string has its ends fixed at x=0 and x = l. At time t = 0 the string is given a shape defined by f(x) = μx(1 − x) where μ is a constant and then released. Find the displacement of the string at any time t'.

ANSWERS