Engineering Mathematics

Running head: ENGINEERING MATHEMATICS
ENGINEERING MATHEMATICS
Name of the Student
Name of the University
Author Note

1ENGINEERING MATHEMATICS
Section 1:
1. The partial differential equation of the wave equation is given by,
( ∂2
∂ t2 )u ( x , t ) – c2∗( ∂2
∂ x2 )u ( x , t ) =0 (1)
Now, let u ( x , t ) can be separated in functions of t and x
u(x,t) = g(x)*f(t) which is a solution of the wave equation.
Hence, equation (1) becomes
g ( x )∗f ’ ’ ( t )=c2∗f ( t )∗g ’ ’ ( x )
 g ( x )
g ’ ’ ( x ) =c2∗f ( t )
f ’ ’ ( t ) (1)
The boundary conditions are the left and the right end are held at height zero.
u(0,t) = 0 for t>0
u(L,t) = 0 for t>0
where, L = length of the string = 0.5 m
Now, the initial displacement is 0 and initial velocity is induced by a sonic wave inducer and
thus it is also 0 as the initial displacement is 0.
Hence, u(0,0) = 0 and du ( 0,0 )
dt = 0
2.
Now, left and right side of the equation (1) is constant let σ.
Hence, g’’(x) – σg(x) = 0. Now, assuming auxiliary solution as exp(rx)

2ENGINEERING MATHEMATICS
( d2
d x2 )exp ( rx ) – σ exp ( rx )=0 (A)
 ( r2 −σ )∗exp ( rx )=0 => r = ±√σ
Also, ( d2
d t2 ) exp ( st ) – c2 exp ( st ) =0 (B)
Assuming exp(st) as the auxiliary solution we get
( s2 – c2 σ ) exp ( st )=0 => s^2 – c^2σ = 0 => s = ±c√σ
Hence, the two independent solution of (A) are exp(c √ σ t ¿ and exp(-c√σ t)
Hence, the general solutions are
X(x) = d1*exp(c √ σ t ¿ + d2*exp(-c√σ t) and T(t) = d3*exp(c√ σ t ¿ + d4*exp(-c√σ t)
Here, d1, d2, d3 and d4 are arbitrary constants.
Now, when σ ≠ 0 then g(x) =0 satisfied when d1 + d2 = 0 => d2 = -d1.
Also, the condition g(L) = 0 is satisfied when
d1*exp( √σ t ¿ + d2*exp(- √ σ t) = 0 => d1(exp(√ σ t ¿−exp(− √ σ t)¿ = 0
Hence, exp(√ σ t ¿−exp(− √ σ t) = 0 => exp(2 √σ L) = 1
Now, there are infinitely many complex numbers that satisfy the above equation and exp(2
√ σ L) = 1 when for any integer k
2√ σ L = 2kπi => √σ = kπi/L => σ = −k2∗π2
L2
Hence, g(x)*f(t) = d1(exp(ikπx/L) – exp(-ikπx/L))(d3exp(ickπt/L) + d4exp(-ickπt/L))
= 2i*d1*sin(kπx/L)[(d3 + d4)cos(ckπt/L) + i(d3 – d4)sin(ckπt/L)]

3ENGINEERING MATHEMATICS
= sin(kπx/L)[αk*cos(ckπt/L) + βksin(ckπt/L)]
Here, αk = 2id1(d3 + d4) and βk = -2d1(d3-d4).
Now, imposing initial condition we get,
u(x,t) = ∑
k =1
∞
sin ( kπx
L ) [αk∗cos ( ckπt
L ) + βk∗sin ( ckπt
L ) ]
The Fourier coefficients needs to satisfy the equations
f(x) = u(x,0) = ∑
k =1
∞
αk∗sin ( kπx
L )
g(x) = du(x,0)/dt = ∑
k =1
∞
βk∗( ckπ
L ) sin ( kπx
L )
Here, the Fourier coefficients are
αk = (2/L) ∫
0
L
f ( x ) sin ( kπx
L ) dx and βk = ( 2
ckπ )∫
0
L
g ( x ) sin ( kπx
L ) dx
3.
The velocity profile is given below,
( ∂
∂ t )u ( x ,0 )= A ( 2 πf 3 ) sin ( 3 πx
L )
A = 1e-3 m, f3 = 540 Hz.
Now, differentiating with respect to t the solution in general form is given by,
u(x , t)=∑
k=1
∞
sin ( kπx
L )[α k cos ( ckπ
L t )+ βk sin ( ckπ
L t )]
Now, the 3rd mode sine wave is given by,