Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Wave Function - Practice Problems | MTH 3102, Assignments of Mathematics

Material Type: Assignment; Professor: Tenali; Class: Intro to Linear Algebra; Subject: Mathematics; University: Florida Institute of Technology; Term: Spring 2009;

Typology: Assignments

Pre 2010

Uploaded on 08/01/2009

koofers-user-pfb
koofers-user-pfb 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Practice Problems - Wave Equation
MTH3102 Spring 2009
(1) Determine the solution of the following initial value problems:
(a) utt c2uxx = 0, u(x, 0) = sin x ut(x, 0) = x2.
(b) utt c2uxx = 0, u(x, 0) = cos x ut(x, 0) = x.
(2) Determine the solution of the initial- boundary value problem:
utt 16uxx = 0,0< x < , t > 0
u(x, 0) = sin x, 0< x <
ut(x, 0) = x2,0< x <
u(0, t)=0, t > 0
(3) Determine the solution of the initial- boundary value problem:
utt 9uxx = 0,0< x < , t > 0
u(x, 0) = 0,0< x <
ut(x, 0) = x3,0< x <
ux(0, t)=0, t > 0
(4) Solve the following initial boundary value problem by method of
seperation of variables:
utt c2uxx = 0,0< x < π, t > 0
u(x, 0) = 3 sin x, 0< x < π
ut(x, 0) = 0,0< x < π
u(0, t) = u(π, t) = 0.
(5) Solve the following initial boundary value problem by method of
seperation of variables:
utt c2uxx = 0,0< x < π, t > 0
u(x, 0) = 0,0< x < π
ut(x, 0) = 8 sin2x, 0< x < π
u(0, t) = u(π, t) = 0.
(6) Solve the following initial boundary value problem by method of
seperation of variables:
utt c2uxx = 0,0< x < π, t > 0
u(x, 0) = sin x, 0< x < π
ut(x, 0) = x2πx, 0< x < π
u(0, t) = u(π, t) = 0.
(7) Solve the following initial boundary value problem by method of
seperation of variables:
utt c2uxx = 0,0< x < π, t > 0
u(x, 0) = cos x, 0< x < π
ut(x, 0) = 0,0< x < π
ux(0, t) = ux(π, t) = 0.
1
pf2

Partial preview of the text

Download Wave Function - Practice Problems | MTH 3102 and more Assignments Mathematics in PDF only on Docsity!

Practice Problems - Wave Equation

MTH3102 Spring 2009

(1) Determine the solution of the following initial value problems: (a) utt − c^2 uxx = 0, u(x, 0) = sin x ut(x, 0) = x^2. (b) utt − c^2 uxx = 0, u(x, 0) = cos x ut(x, 0) = x. (2) Determine the solution of the initial- boundary value problem: utt − 16 uxx = 0 , 0 < x < ∞, t > 0 u(x, 0) = sin x, 0 < x < ∞ ut(x, 0) = x^2 , 0 < x < ∞ u(0, t) = 0 , t > 0 (3) Determine the solution of the initial- boundary value problem: utt − 9 uxx = 0 , 0 < x < ∞, t > 0 u(x, 0) = 0 , 0 < x < ∞ ut(x, 0) = x^3 , 0 < x < ∞ ux(0, t) = 0 , t > 0 (4) Solve the following initial boundary value problem by method of seperation of variables: utt − c^2 uxx = 0 , 0 < x < π, t > 0 u(x, 0) = 3 sin x, 0 < x < π ut(x, 0) = 0 , 0 < x < π u(0, t) = u(π, t) = 0. (5) Solve the following initial boundary value problem by method of seperation of variables: utt − c^2 uxx = 0 , 0 < x < π, t > 0 u(x, 0) = 0 , 0 < x < π ut(x, 0) = 8 sin^2 x, 0 < x < π u(0, t) = u(π, t) = 0. (6) Solve the following initial boundary value problem by method of seperation of variables: utt − c^2 uxx = 0 , 0 < x < π, t > 0 u(x, 0) = sin x, 0 < x < π ut(x, 0) = x^2 − πx, 0 < x < π u(0, t) = u(π, t) = 0. (7) Solve the following initial boundary value problem by method of seperation of variables: utt − c^2 uxx = 0 , 0 < x < π, t > 0 u(x, 0) = cos x, 0 < x < π ut(x, 0) = 0 , 0 < x < π ux(0, t) = ux(π, t) = 0. 1

2

(8) A string is streched between the fixed points 0, and 1 on the x axis, and released at rest from the position y = A sin πx, where A is a constant. Obtain an expression for the subsequent displacements y(x, t). (9) A string streched between the points 0 and π on the x axis and initially at rest, is released from position y = f (x).Its motion is opposed by resistence, which is proportional to the velocity at each point. Let the unit of time chosen so that the equation of motion becomes ytt = yxx − 2 βyt, 0 < x < π, t > 0 where β is a positive constant. Assuming, 0 < β < 1 derive the expression

y(x, t) = e−βt

∑^ ∞

n=

Bn

cos αnt +

β αn

sin αnt

sin nx

where αn =

n^2 − β^2 , Bn = (^2) π

∫ (^) π 0 f^ (x) sin^ nx dx, n^ = 1,^2 ,^ · · ·^. (10) The ends of a streched string are fixed at the origin and at the point x = π on the horizontal x axis. The string is initially at rest along the x axis and then drops under its own weight. The vertical displacements y(x, t) satisfies the DE ytt = a^2 yxx − g, 0 < x < π, t > 0 where g is accelaration due to gravity. Use the method of variation of paramaters to find the dispalcements.