2. Submitted To:
Ms.Binish Farhan
Submitted By:
Iqra bibi 59
Shehar bano 62
Saman israr 124
Haleema urooj 123

3. Give a detailed analysis of wave equation
regarding its formulation solution and
interpretation and also give an expression for
forced vibrating membranes

4. INTRODUCTION
 The scope of our discussions, we investigate the vibrations of perfectly elastic
strings and membranes.
 We begin by formulating the governing equations for a vibrating string from
physical principles.
 The appropriate boundary conditions will be shown to be similar in a mathematical
sense to those boundary conditions for the heat equations.

5. Wave Equation:
The wave equation is an important second-order linear partial
differential equation for the description of waves—as they occur in
classical physics—such as mechanical waves or light waves,
sound wave. It arises in fields like acoustics,
electromagnetic, and fluid dynamics.

6. History:
Historically, the problem of a vibrating string such as that of a musical instrument was
studied by Jean le Rond d'Alembert, Leonhard Euler, Daniel Bernoulli, and Joseph-
Louis Lagrange. In 1746, d’Alembert discovered the one-dimensional wave
equation, and within ten years Euler discovered the three-dimensional wave
equation.

7. Physical interpretation:
 A solution of this equation can be quite complicated, but it can be analyzed as a
linear combination of simple solutions that are sinusoidal plane waves with various
directions of propagation and wavelengths but all with the same propagation speed
c. This analysis is possible because the wave equation is linear; so that any multiple
of a solution is also a solution, and the sum of any two solutions is again a solution.
This property is called the superposition principle in physics.
 The wave equation is the simplest example of a hyperbolic differential equation. It,
and its modifications, plays fundamental roles in continuum mechanics, quantum
mechanics, plasma physics, general relativity, geophysics, and many other scientific
and technical disciplines

8. Cont…
 The wave equation alone does not specify a physical solution; a unique solution is
usually obtained by setting a problem with further conditions, such as initial
conditions, which prescribe the amplitude and phase of the wave. Another important
class of problems occurs in enclosed spaces specified by boundary conditions, for
which the solutions represent standing waves, or harmonics, analogous to the
harmonic of musical instruments.

9. DERIVATION OF A VERTICALLY VIBRATING
STRING
 A string vibrates only if it is tightly stretched. Consider a horizontally stretched string in its
equilibrium configuration,
 We imagine that the ends are tied down in some way maintaining the tightly stretched
nature of the string. You may wish to think of stringed musical instruments as examples.
 We begin by tracking the motion of each particle that comprises the string. We let α be the
x-coordinate of a particle when the string is in the horizontal equilibrium position.
 The string moves in time; it is located somewhere other than the equilibrium position at
time t.
 The trajectory of particle α is indicated with both horizontal and vertical components.

10. Cont…
 We will assume the slope of the string is small, in which case it can be shown that the
horizontal displacement v can be neglected.
 As an approximation, the motion is entirely vertical, x = α. In this situation, the vertical
displacement u depends on x and t:
 y = u(x,t). (1)
 In general( ), it is best to let y = u(α,t).
Vertical and horizontal displacements of a particle on a highly stretched string.
ax 

11. Newton’s law
 . We consider an infinitesimally thin segment of the string contained between x and
 x +Δ x
 In the unperturbed (yet stretched) horizontal position, we assume that the mass
density is known.
 For the thin segment, the total mass is approximately (x)Δ x. Our object is to derive
a partial differential equation describing how the displacement u changes in time.
 Accelerations are due to forces; we must use Newton’s law. For simplicity we will
analyze Newton’s law for a point mass:
F = ma (2)
)(0 xp
0p

12.  We must discuss the forces acting on this segment of the string. There are body forces,
which we assume act only in the vertical direction (e.g., the gravitational force), as
well as forces acting on the ends of the segment of string.
 We assume that the string is perfectly flexible, it offers no resistance to bending. This
means that the force exerted by the rest of the string on the endpoints of the segment of
the string is in the direction tangent to the string.
 This tangential force is known as the tension in the string, and we denote its magnitude
by T(x,t).
 In Fig.2 we show that the force due to the tension (exerted by the rest of the string)
pulls at both ends in the direction of the tangent, trying to stretch the small segment.
 To obtain components of the tensile force, the angle θ between the horizon and the
string is introduced. The angle depends on both the position x and time t.
 Furthermore, the slope of the string may be represented as either dy/dx or tanθ:
 slope = dy/dx= tanθ(x,t)=∂u/∂x (3)

13.  FIGURE .2
Stretching of a finite segment of string, illustrating the tensile forces.

14.  The horizontal component of Newton’s law prescribes the horizontal motion, which we claim is small
and can be neglected.
 The vertical equation of motion states that the mass (x)Δ x times the vertical component of
acceleration ( , where∂/∂t is used since x is fixed for this motion) equals the vertical
component of the tensile forces plus the vertical component of the body forces:
(4)
 Where T(x,t) is the magnitude of the tensile force and Q(x,t) is the vertical component of the body
force per unit mass.
 Dividing (4) by δx and taking the limit as δx →0 yields
(5)
for small angles θ,
and hence (5) becomes (6)
0p
tu 22
/
),()(][)( 02
2
0 txQxp
x
u
T
xt
u
xp 







),()()],(sin),([)( 02
2
0 txQxptxtxT
xt
u
xp 









 sin
cos
sin
tan 


x
u

15. Perfectly elastic strings
 . The tension of a string is determined by experiments. Real strings are nearly perfectly
elastic, by which we mean that the magnitude of the tensile force T(x,t) depends only
on the local stretching of the string.
 Since the angle θ is assumed to be small, the stretching of the string is nearly the same
as for the unperturbed highly stretched horizontal string, where the tension is constant,
T0 (to be in equilibrium). Thus, the tension T(x,t) may be approximated by a constant .
 Consequently, the small vertical vibrations of a highly stretched string are governed by
(7)
 One-dimensional wave equation. If the only body force per unit mass is gravity, then
Q(x,t)=−g in (7).
 In many such situations, this force is small
(relative to the tensile force ) and can be neglected.
0T
)(),()( 02
2
02
2
0 xtxQ
x
u
T
t
u
x  





22
00 / xuTg 

16.  Alternatively, gravity sags the string, and we can calculate the vibrations with respect to
the sagged equilibrium position.
 . In either way we are often led to investigate (7) in the case in which Q(x,t) = 0,
 (8), (9)
where
Equation (9) is called the one-dimensional wave equation.
The notation is introduced because has the dimension of velocity squared.
We will show that c is a very important velocity. For a uniform string, c is constant
2
2
02
2
0 )(
x
u
T
t
u
x





 2
2
2
2
2
x
u
c
t
u





0
02

T
c 
2
c
)(0
0
x
T


17. VIBRATING MEMBRANE
 The vibration of a string (one dimension) can be extended to the vibration of a
membrane (two dimensions).
 The vertical displacement of a vibrating string satisfies the one-dimensional wave
equation
 There are important physical problems that solve
(4.1)
 known as the two-dimensional wave equation
2
2
2
2
2
x
u
c
t
u





)( 2
2
2
2
22
2
2
y
u
x
u
uc
t
u









18.  An example of a physical problem that satisfies a two-dimensional wave equation is
the vibration of a highly-stretched membrane. This can be thought of as a two-
dimensional vibrating string.
 We again introduce the displacement z = u(x,y,t), which depends on x,y, and t
 If all slopes (i.e., ∂u/∂x and ∂u/∂y) are small, then, as an approximation, we may
assume that the vibrations are entirely vertical and the tension is approximately
constant.
 Then the mass density (mass per unit surface area), , of the membrane in the
unperturbed position does not change appreciably when the membrane is perturbed.
 We assume all slopes (∂u/ ∂x and ∂u /∂y ) are small, so that, as an approximation,
the vibrations are entirely vertical and the tension is approximately constant.
),(0 yx

19.  We consider any small rectangular segment of the membrane between x and and
y and
 Here, is the constant force per unit length exerted by the stretched membrane on any
edge of this section of the membrane.
 Newton’s law for the vertical motion of the small segment at (x, y) is
(4.2)
 where F is the sum of all the vertical forces and m is the mass of this section of the
membrane:
where is the mass density of the membrane (mass per unit area).
Fzyx
t
u
m 


),,(2
2
yxm  0
0
xx 
yy 
oT

20.  FIGURE 4.1
 Slightly stretched membrane with approximately constant tension oT

21.  The vertical forces are the vertical components of the tensile forces in effect along all
four sides.
 For example, on the side where x is fixed at , y varies from y to , the
vertical component of tension (per unit length) is sinφ, where φ is the angle of
inclination for fixed y.
 Since φ is assumed small, sinφ ≈ tanφ = ∂u/∂x, where this is to be evaluated at each
value of y. To determine the entire contribution at this one edge, we must sum up
(integrate) from y to y +Δ y for fixed x.
 However, φ will not vary by much, and we claim that we may approximate sinφ by
∂u/∂x(x +Δx,y) for all y from y to y +Δ y.
 Thus, the total contribution to the vertical component of tension for this one side is
obtained by multiplying the value at each point by the length of the side
Δy:
yy 
),(0 yxx
x
u
T 


xx 

22.  Furthermore, we should check the sign.
 If ∂u/∂x > 0 then the vertical component of the tensile force is upward at x+Δx (but
downward at x).
 In a similar way, we can analyze the vertical components along the other sides. Thus
(4.3)
 Using the Taylor series (linearization),
and
where f = ∂u/∂x and g = ∂u/∂y,
)],(),(),(),([),,( 02
2
0 yx
y
u
xyyx
y
u
xyx
x
u
yyxx
x
u
yTtyx
t
u
yx















y
g
xyxyyxf


 ),(),(
y
g
xyxgyyxg


 ),(),(

23.  we obtain
(4.4)
 Dividing (4.4) by and taking the limit as both Δx →0 and Δ y →0 yields
the two-dimensional wave equation
(4.5)
 where again
2
2
2
2
002
2
0 ),,(
y
u
x
u
yxT
y
u
yx
u
x
yxTtyx
t
u
yx


















yx0
2
2
2
2
2
2
2
y
u
x
u
c
t
u








0
02

T
c 