Question 7.3: When solving the determinant equation (7.19), we have made a...

When solving the determinant equation (7.19), we have made a mathematical restriction for c by setting c = 2  cos Θ.
Show that for the cases

(a) |c| = 2,
(b) c <−2

he eigenvalue equation DND_{N} = 0 cannot be satisfied. Clarify that thereby the special choice of the constant c is justified.

DN=cDN1DN2,  if    N2.D_{N} = cD_{N−1} −D_{N−2},   if        N ≥ 2.                               (7.19)

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

(a)

Dn=cDn1Dn2,           D1=c=±2,      D0=1.D_{n} = cD_{n−1} −D_{n−2},            D_{1} = c=±2,            D_{0} = 1.                                (7.31)

We assert and prove by induction

DnDn1.|D_{n}| ≥ |D_{n−1}|.                                (7.32)

Induction start: n=2,D0=1,D1=2,D2=3.n = 2, |D_{0}| = 1, |D_{1}| = 2, |D_{2}| = 3.

Induction conclusion from n −1, n−2 to n:

Dn2=4Dn12±4Dn1Dn2+Dn22|D_{n}|^{2} = 4|D_{n−1}|^{2} ±4|D_{n−1}||D_{n−2}| + |D_{n−2}|^{2}

4Dn12+Dn224Dn1Dn2≥ 4|D_{n−1}|^{2} + |D_{n−2}|^{2} − 4|D_{n−1}||D_{n−2}|

Dn2Dn123Dn12+Dn224Dn1Dn2.⇒ |D_{n}|^{2} − |D_{n−1}|^{2} ≥ 3|D_{n−1}|^{2} + |D_{n−2}|^{2} − 4|D_{n−1}||D_{n−2}|.

According to the induction condition,

Dn1=Dn2+ϵ|D_{n−1}| = |D_{n−2}| + \epsilon                           with ϵ\epsilon  ≥ 0.

From this, it follows that

Dn2Dn124Dn22+6ϵDn2+3ϵ24ϵDn24Dn22|D_{n}|^{2} − |D_{n−1}|^{2} ≥ 4|D_{n−2}|^{2} +6\epsilon|D_{n−2}| + 3\epsilon^{2} −4\epsilon|D_{n−2}| − 4|D_{n−2}|^{2}

2ϵDn2≥ 2\epsilon|D_{n−2}|

0≥ 0

DnDn1.⇒ |D_{n}| ≥ |D_{n−1}|.                                   (7.33)

Since Dn|D_{n}| monotonically increases in n, and D1=2>0|D_{1}| = 2 > 0, we have DN>0|D_{N}| > 0. Therefore DN=0D_{N} = 0 cannot be satisfied. ω=0 and  ω=2T/maω = 0   and   ω =\sqrt{2T/ma} are not eigenfrequencies of the vibrating chain.

(b) By inserting the ansatz Dn=Apn,p0D_{n} = Ap^{n}, p ≠ 0, we also find the solution of the recursion formula Dn=cDn1Dn2,D1=c,D0=1D_{n} = cD_{n−1} −D_{n−2}, D_{1} = c, D_{0} = 1:

  p1=12(c+(c24)1/2)<0p2=12(c(c24)1/2)<0}0>p1>p2   \left. \begin{matrix} p_{1} = \frac{1}{2} \left(c +(c^{2} −4)^{1/2}\right)< 0 \\ p_{2} = \frac{1}{2} \left(c -(c^{2} −4)^{1/2}\right)< 0 \end{matrix} \right\} \quad \quad 0>p_{1}>p_{2}                                         (7.34)

The general solution for incorporating the boundary conditions D0=1,D1=cD_{0} = 1, D_{1} = c reads

Dn=A1p1n+A2p2n.D_{n} = A_{1}p^{n}_{1} + A_{2}p^{n}_{2} .                                                  (7.35)

With D0=1,D1=cD_{0} = 1, D_{1} = c, it follows that

A1+A2=1A_{1} +A_{2} = 1,

A12(c+(c24)1/2)+A22(c(c24)1/2)=c,\frac{A_{1}}{2} \left(c + (c^{2} −4)^{1/2} \right)+ \frac{A_{2}}{2} \left(c − (c^{2} −4)^{1/2}\right)= c,

 

A1=c+(c24)1/22(c24)1/2A2=c+(c24)1/22(c24)1/2.A_{1} =\frac{ c +(c^{2} −4)^{1/2}}{2(c^{2} − 4)^{1/2}}⇔ A_{2} =\frac{−c +(c^{2} −4)^{1/2}}{2(c^{2} − 4)^{1/2}} .                                       (7.36)

One then has

Dn=12c+(c24)1/2(c24)1/2p1n+12(c24)1/2c(c24)1/2p2nD_{n} = \frac{1}{2} \frac{c +(c^{2} −4)^{1/2}}{(c^{2} −4)^{1/2}} p^{n}_{1}+ \frac{1}{2} \frac{(c^{2} −4)^{1/2} −c}{(c^{2} −4)^{1/2}} p^{n}_{2}

 

=1(c24)1/2(p1n+1p2n+1).= \frac{1}{(c^{2} − 4)^{1/2}}\left(p^{n+1}_{1}−p^{n+1}_{2}\right) .                              (7.37)

To determine the physically possible vibration modes, we had required that DN=0D_{N} = 0:

DN=0(p2p1)N+1=1.D_{N}= 0 ⇒ \left(\frac{p_{2}}{p_{1}}\right)^{N+1}= 1.                                  (7.38)

But now 0>p1>p20 > p_{1} > p_{2}, hence (p2/p1)N+1>1(p_{2}/p_{1})^{N+1} > 1. Thus, for the case c < −2 eigenfrequencies do not exist too.

These supplementary investigations can be summarized as follows: The possible eigenfrequencies of the vibrating chain lie between 0 and 2T/ma\sqrt{2T/ma}:

0<ω<2Tma.0 < |ω| <\sqrt{\frac{2T}{ma}}.                                            (7.39)

Related Answered Questions