Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 7 / b

Table of contents
- Solution b

Solution

Substitution of

μ = 2

into the normal modes of three masses connected by two equal springs gives

Ω = \frac{k}{m}, ω_{0} = 0, ω_{1} = Ω, ω_{2} = 2 Ω.

The corresponding shapes are

v_{0} = (1, 1, 1)^{T}

,

v_{1} = (1, 0, - 1)^{T}

,

v_{2} = (1, - 1, 1)^{T}

. For example,

K v_{2} = k (2, - 4, 2)^{T} = 2 (k / m) M v_{2}

, verifying the third eigenvalue directly.

Their relevant orthogonality uses the kinetic matrix

M = m diag (1, 2, 1)

:

v_{i}^{T} M v_{j} = 0 (i \neq = j), (v_{0} v_{1} v_{2})^{T} M (v_{0} v_{1} v_{2}) = m diag (4, 2, 4) .

This is mass-matrix orthogonality of normal modes. It also follows by subtracting

v_{i}^{T} K v_{j} = ω_{j}^{2} v_{i}^{T} M v_{j}

from its transpose and using the symmetry of

K, M

. Ordinary Euclidean inner products need not vanish: in fact

v_{0}^{T} v_{2} = 1

.

Figure 1.
The three normal modes for a free spring chain with middle mass $2 m$ ; amplitudes are relative and the zero mode is translation
. Details.

 Ancestors (12)

 View article source

 Discussion (0)

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