This set of Partial Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Derivation of One-dimensional Wave Equation”.
1. Consider one dimensional wave equation utt = 4 uxx; 0 ≤ x ≤ π, t ≥ 0 satisfying the boundary conditions u (0,t) = 0 and u(π,t) = 0 and initial conditions u(x,0) = 2sin(3x) – sinx and ut(x,0) = 0. What is u (π, π)?
a) -1
b) -π
c) 0
d) -3
View Answer
Explanation: The general solution is given by u(x, t) = ∑An Sin (nx) cos (2nt)
To find coefficients:
The initial conditions are given as u(x, 0) = ∑ An Sin (nx) = 2sin (3x) – sinx.
So, A1 = 2, A3 = – 1 and all other An = 0.
Therefore,
U (x, t) = 2 sin(x) cos (2t) − sin (3x) cos (6t)
U (π, π) = 2 sin (π) cos (2π) – sin (3π) cos (6π) = 0.
2. By which of the following methods can the one dimensional equation be solved?
a) Separation of variables
b) Cauchy’s equation
c) Schrodinger’s wave equation
d) Euler’s equation
View Answer
Explanation: The one dimensional wave equation can be solved by separation of variables.
This has a solution ψ (x, t) = X (x) T (t)
Where, the solution of X is X (x) = C cos (kx) + D sin (kx) and
The solution of T is T (t) = E cos (ωt) + F sin (ωt).
3. Which of the following is a possible solution of one dimensional wave equation?
a) Y(x, t) = (Ax + B) (Ct + D)
b) Y(x, t) = (Ax + Bt)
c) Y(x, t) = A eαt
d) Y(x, t) = A sin αt
View Answer
Explanation: The possible solutions of one dimensional wave equation are
Y (x, t) = (Ax + B) (Ct + D)
Y (x, t) = (A cosλx + B sinλx) (C cosλat + D sinλat)
Y (x, t) = (A eλx + B e-λx) (C eλat + D e-λat)
4. Which of the following is considered to be the suitable solution of one dimensional wave equation?
a) Y(x, t) = (A cosλx + B sinλx) (C cosλat + D sinλat)
b) Y(x, t) = A eλx + B e-λx
c) Y(x, t) = A sinλx + B cosλx
d) Y(x, t) = (Ax + Bt)
View Answer
Explanation: The most suitable solution of one dimensional wave equation is
Y(x, t) = (A cosλx + B sinλx) (C cosλat + D sinλat). This solution is due to the linearity of utt = α2 uxx.
This has a boundary condition from (-∞, ∞).
5. Which of the following is considered to be d’Alembert’s solution?
a) U(x, t) = \(\frac {1}{2}\) {Φ(x + t) + Φ(x – t)}
b) U(x, 0) = 0
c) U(x, t) = Φ(x + t)
d) Ut(x, 0) = ψ(x)
View Answer
Explanation: d’Alembert devised his solution in 1746, and Euler subsequently expanded the method in 1748. The solution has a formula (x, t) = \(\frac {1}{2}\) {Φ(x + t) + Φ(x – t)}. For a given initial displacement and the velocity in the vertical direction, the wave equation is completely solved and this solution is called the progressive wave solution.
6. The equation is parabolic if B2 – 4AC = 0, which is the reduced canonical form of second order linear Partial Differential Equation.
a) False
b) True
View Answer
Explanation: Any equation of second order semi linear Partial Differential Equation can be reduced to one of the following canonical forms:
- Elliptic equation if B2 – 4AC < 0
- Parabolic equation if B2 – 4AC = 0
- Hyperbolic equation if B2 – 4AC > 0
7. Classify the partial differential equation 4uxx = ut.
a) Elliptic
b) Hyperbolic
c) Parabolic
d) Spherical
View Answer
Explanation: Given that 4uxx – ut = 0
A = 4, B = 0, C = 0
B2 – 4AC = (0) – (4 X 4 X 0) = 0
Therefore, the given partial differential equation is parabolic.
8. Which of the following is the boundary condition in which the initial position of string is f(x) and the initial velocity imparted at each point x is g(x)?
a) Ut(x, 0) = g(x)
b) U(x, 0) = 0
c) U (0, t) = t
d) U (0, 0) = x
View Answer
Explanation: The boundary conditions of one dimensional wave equation are
- u (0,t) = 0
- u(x, 0) = f(x)
- u(l, t) = 0
- ut(x,0) = g(x)
9. The one dimensional wave equation is non periodic in nature.
a) False
b) True
View Answer
Explanation: The one dimensional wave equation is periodic in nature whereas the one dimensional heat equation is non periodic in nature. The period of the wave is the time for a particle on a medium to make one complete vibration cycle.
10. In the wave equation Utt = c2 Uxx, what does c2 stand for?
a) Volume/density per m3
b) Tension/mass per unit length
c) Mass/length per KN
d) Newton/second per unit length
View Answer
Explanation: C2 = T/m, Tension per Unit Length, which is determined by wave velocity equation of string. The wave velocity is given by
v = mass of string/ length of string = m/l.
Generally, the speed of the wave can be found from linear density and tension.
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