The Chemistry Maths Book, Second Edition

366 Chapter 12Second-order differential equations. Constant coefficients

20. 
    

21.Solve subject to the conditionθ(t 1 + 12 πτ) 1 = 1 θ(t).

Section 12.5

22.Given that the general solution of the equation of motionmP1= 1 −kxfor the harmonic

oscillator isx(t) 1 = 1 a 1 cos 1 ωt 1 + 1 b 1 sin 1 ωt, where , (i)show that the solution can be

written in the formx(t) 1 = 1 A 1 cos(ωt 1 − 1 δ), where Ais the amplitude of the vibration and δ

is the phase angle, and express Aand δin terms of aand b; (ii)find the amplitude and

phase angle for the initial conditionsx(0) 1 = 11 , H(0) 1 = 1 ω.

23.Solve the equation of motion for the harmonic oscillator with initial conditionsx(0) 1 = 10 ,

H(0) 1 = 1 u

0

.

Section 12.6

24.For the particle in a box, find the nodes and sketch the graph of the wave functionψ

n

for

(i)n 1 = 14 and (ii)n 1 = 15.

25. (i) Solve the Schrödinger equation (12.44) for the particle in a box of length lwith

potential-energy functionV 1 = 10 for−l 221 ≤ 1 x 1 ≤ 1 +l 22 , V 1 = 1 ∞forx 1 ≤ 1 −l 22 andx 1 ≥ 1 +l 22.

(ii) Show that the solutionsψ

n

are even functions of xwhen nis odd and odd functions

when nis even. (iii) Show that the solutions are the same as those given by (12.53) if xis

replaced byx 1 + 1 l 22 , except for a possible change of sign.

26.For the particle in the box in Section 12.6, show that wave functions

forn 1 = 11 andn 1 = 12 are (i)normalized, (ii)orthogonal.

Section 12.7

27.For the particle in a ring show that wave functions forn 1 = 13 and

n 1 = 14 are (i)normalized, (ii)orthogonal.

28.The diagrams of Figure 12.8 are maps of the signs and nodes of some real wave functions

(12.71) for the particle in a ring. Draw the corresponding diagrams for (i) n 1 = 1 ± 3 ,

(ii)n 1 = 1 ± 4.

29.Verify that equation (12.72) and its solutions (12.74) are transformed into (12.62) and

(12.65) by means of the change of variableθ 1 = 1 x 2 r.

Section 12.8

30.Find a particular solution of the differential equationy′′ 1 − 1 y′ 1 − 16 y 1 = 121 + 13 x.

Find the general solutions of the differential equations:

31.y′′ 1 − 1 y′ 1 − 16 y 1 = 121 + 13 x 32.y′′ 1 − 18 y′ 1 + 116 y 1 = 111 − 14 x

3

33.y′′ 1 − 1 y′ 1 − 16 y 1 = 12 e

− 3 x

34.y′′ 1 − 1 y′ 1 − 12 y 1 = 13 e

−x

35.y′′ 1 − 18 y′ 1 + 116 y 1 = 1 e

4 x

36.y′′ 1 − 1 y′ 1 − 16 y 1 = 121 cos 13 x

37.y′′ 1 + 14 y 1 = 131 sin 12 x 38.y′′ 1 − 1 y′ 1 − 16 y 1 = 121 + 13 x 1 + 12 e

− 3 x

1 + 121 cos 13 x

Section 12.9

39.An RLC-circuit contains a resistor (resistance R), an inductor (inductance L), and a

capacitor (capacitance C) connected in series with a source of e.m.f. E.

ψ θ

θ

n

in

()= 12 πe

ψ

n

x

l

nx

l

() sin=

2 π

ω= km

d

dt

a

2

2

2

0

θ

+=θ

dy

dx

dy

dx

yy y x

2

2

+−=;20 02 0()=,→ as →∞