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 DN = 0 cannot be satisfied. Clarify that thereby the special choice of the constant c is justified.
DN=cDN−1−DN−2, if N≥2. (7.19)
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(a)
Dn=cDn−1−Dn−2, D1=c=±2, D0=1. (7.31)
We assert and prove by induction
∣Dn∣≥∣Dn−1∣. (7.32)
Induction start: n=2,∣D0∣=1,∣D1∣=2,∣D2∣=3.
Induction conclusion from n −1, n−2 to n:
∣Dn∣2=4∣Dn−1∣2±4∣Dn−1∣∣Dn−2∣+∣Dn−2∣2
≥4∣Dn−1∣2+∣Dn−2∣2−4∣Dn−1∣∣Dn−2∣
⇒∣Dn∣2−∣Dn−1∣2≥3∣Dn−1∣2+∣Dn−2∣2−4∣Dn−1∣∣Dn−2∣.
According to the induction condition,
∣Dn−1∣=∣Dn−2∣+ϵ with ϵ ≥ 0.
From this, it follows that
∣Dn∣2−∣Dn−1∣2≥4∣Dn−2∣2+6ϵ∣Dn−2∣+3ϵ2−4ϵ∣Dn−2∣−4∣Dn−2∣2
≥2ϵ∣Dn−2∣
≥0
⇒∣Dn∣≥∣Dn−1∣. (7.33)
Since ∣Dn∣ monotonically increases in n, and ∣D1∣=2>0, we have ∣DN∣>0. Therefore DN=0 cannot be satisfied. ω=0 and ω=2T/ma are not eigenfrequencies of the vibrating chain.
(b) By inserting the ansatz Dn=Apn,p=0, we also find the solution of the recursion formula Dn=cDn−1−Dn−2,D1=c,D0=1:
p1=21(c+(c2−4)1/2)<0p2=21(c−(c2−4)1/2)<0}0>p1>p2 (7.34)
The general solution for incorporating the boundary conditions D0=1,D1=c reads
Dn=A1p1n+A2p2n. (7.35)
With D0=1,D1=c, it follows that
A1+A2=1,
2A1(c+(c2−4)1/2)+2A2(c−(c2−4)1/2)=c,
A1=2(c2−4)1/2c+(c2−4)1/2⇔A2=2(c2−4)1/2−c+(c2−4)1/2. (7.36)
One then has
Dn=21(c2−4)1/2c+(c2−4)1/2p1n+21(c2−4)1/2(c2−4)1/2−cp2n
=(c2−4)1/21(p1n+1−p2n+1). (7.37)
To determine the physically possible vibration modes, we had required that DN=0:
DN=0⇒(p1p2)N+1=1. (7.38)
But now 0>p1>p2, hence (p2/p1)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:
0<∣ω∣<ma2T. (7.39)