Linearly damped string

With fixed endpoints,

y_{tt} - y_{xx} + y_{t} = 0

separates into sine modes whose amplitudes satisfy

T_{n}^{''} + T_{n}^{'} + n^{2} π^{2} T_{n} = 0.

(1)

Table of contents

Critical damping

 0  0

Critical damping occurs when a second-order mode has a repeated real characteristic root. It returns to equilibrium without oscillating and as quickly as possible among nonoscillatory regimes.

Separated solution of the damped string equation

 0  0

For initial sine coefficient

a_{n}

and zero initial velocity, put

ω_{n}^{2} = n^{2} π^{2} - 1/4

. The modal amplitude is

T_{n} (t) = a_{n} e^{- t /2} (cos (ω_{n} t) + \frac{s i n ( ω _{n} t )}{2 ω _{n}}) .

(1)

Energy dissipation identity for a linearly damped string

 0  0

For fixed endpoints and

E (t) = \frac{1}{2} \int_{0}^{1} (y_{t}^{2} + y_{x}^{2}) d x,

(1)

integration by parts and the damped wave equation give

E^{'} (t) = - \int_{0}^{1} y_{t}^{2} d x \leq 0.

(2)

 Ancestors (7)

 Incoming links (1)

Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1 / 13B / a / Solution

 View article source

 Discussion (0)

New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version

Linearly damped string

Critical damping

Separated solution of the damped string equation

Energy dissipation identity for a linearly damped string

 Ancestors (7)

 Incoming links (1)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)