4.1
The kinetic energy of the system is the sum of the separate kinetic energy of the two masses, i.e.:
The potential energy of the system is the sum of the separate potential energy of the two masses, i.e.: Noting that:
m X
which are the expressions given by (4.4a) and (4.4b) 4.2
The total energy of Problem 4.1 can be written as:
2 2
2 2
2 ( ) ( )
2 1 2
1 2
1 x y s x y
l y mg m x
m E
E
E = k+ p = & + & + + + −
The above equation can be rearranged as the format:
xy pot y pot kin x pot
kin E E E E
E
E =( + ) +( + ) +( )
where, 2 2
2 2
) 1
( s x
l x mg m E
Ekin pot x ⎟
⎠
⎜ ⎞
⎝
⎛ +
+
=
+ & , 2 2
2 2
) 1
( s y
l y mg m E
Ekin pot y ⎟
⎠
⎜ ⎞
⎝
⎛ +
+
=
+ & ,
and Epot =−2sxy
4.3 a
x=−2 , y=0:
=0
x , y=−2a:
4.4
For mass m , Newton’s second law gives: 1 sx x m1&&1 = For mass m , Newton’s second law gives: 2
sx x
m2 &&2 =−
≡
x=0 -2a -a -a
+
-a a
Y
--X y=0
≡
-2a -a -a
+
-a a
Y +
-X
Provided x is the extension of the spring and l is the natural length of the spring,
which shows the system oscillate at a frequency:
ω2 = μs
For a sodium chloride molecule the interatomic force constant s is given by:
]
If the upper mass oscillate with a displacement of x and the lower mass oscillate with a displacement of y, the equations of motion of the two masses are given by Newton’s second law as:
)
Suppose the system starts from rest and oscillates in only one of its normal modes of frequency ω, we may assume the solutions:
t
Using these solutions, the equations of motion become:
We may, by dividing through by meiωt, rewrite the above equations in matrix form as:
which has a non-zero solution if and only if the determinant of the matrix vanishes;
that is, if into equation (4.5.1), we have:
2
which is the ratio of the amplitude of the upper mass to that of the lower mass.
Similarly, in the slower mode, ω2 =(3+ 5)s 2m. By substitution of the value of frequency into equation (4.5.1), we have:
2
The motions of the two pendulums in Figure 4.3 are given by:
t
Supposing the spring is very weak, the stiffness of the spring is ignorable, i.e. s≈0.
Noting that ω12 =g l and ω22 =(g l+2s m), we have:
Hence, the energies of the masses are given by:
( )
The total energy is given by:
2
which show that the constant energy E is completed exchanged between the two pendulums at the beat frequency (ω2−ω1).
4.7
By adding up the two equations of motion, we have:
)
+ &&
&& m && &&
which can be written as:
X&&+ω12X =0 (4.7.1)
On the other hand, the equations of motion can be written as:
) (
) (
2 1
y m x
y s l y g
y m x x s l x g
− +
−
=
−
−
−
=
&&
&&
By subtracting the above equations, we have:
) ( )
(
2 1
y m x
s m y s l x y g
x ⎟⎟⎠ −
⎜⎜ ⎞
⎝
⎛ +
−
−
−
− &&=
&&
i.e. 1 1 ( ) 0
2 1
=
⎥ −
⎦
⎢ ⎤
⎣
⎡ ⎟⎟⎠
⎜⎜ ⎞
⎝
⎛ +
+ +
− x y
m s m
l y g x&& &&
which can be written as:
2 0
2 =
+ Y
Y&& ω (4.7.2) where,
y x
Y = − and ⎟⎟⎠
⎜⎜ ⎞
⎝
⎛ +
+
=
2 1 2
2
1 1
m s m
l ω g
Equations (4.7.1) and (4.7.2) take the form of linear differential equations with constant coefficients and each equation contains only one dependant variable, therefore X and Y are normal coordinates and their normal frequencies are given by ω1 and ω2 respectively.
4.8
Since the initial condition gives x& = y& =0, we may write, in normal coordinate, the solutions to the equations of motion of Problem 4.7 as:
t Y
Y
t X
X
2 0
1 0
cos cos
ω ω
=
=
i.e.
t Y
y x
t m X
m y m x m
2 0
1 0 2
1 2 1
cos cos ω
ω
=
− + = +
By substitution of initial conditions: t=0, x= and A y=0 into the above equations, we have:
A Y
A M m X
=
=
0
1
0 ( )
where, M =m1+m2
The solutions to the above equations are given by:
)
From the analysis in Problem 4.6, we know, at weak coupling conditions, cosωmt and sinωmt are constants over one cycle, and the relation: ωa ≈ g l, so the energy of the mass m1, E , and the energy of the mass x m2, E , are the sums of their y separate kinetic and potential energies, i.e.:
2
By substitution of the expressions of x and y in terms of cosωat and sinωat given by Problem 4.8 into the above equations, we have:
[ ]
Add up the two equations and we have:
t Subtract the two equations and we have:
t
i.e. s Y F t Equations (4.10.1) and (4.10.2) shows that the normal coordinates X and Y are those for damped oscillators driven by a force F0cosωt.
By neglecting the effect of r , equation of (4.10.1) and (4.10.2) become:
t substitution of the solutions to the above equations, we have:
t
These equations satisfy any t if
0
⎥⎦ and the tension of the spring is T . Equations of motion give:
Mx&&+kx=F0cosωt+T (4.11.1)
Eliminating T , we have:
) (
0cos t s y x
F kx x
M&&+ = ω + − so for x=0 at all times, we have
0
0cos t+ sy=
F ω
that is
s t y=−F0 cosω Equation (4.11.2) and (4.11.3) now give:
=0 + sy y m&&
with ω2 =s m, so M is stationary at ω2 =s m.
This value of ω satisfies all equations of motion for x=0 including t
F T =− 0cosω
4.12
Noting the relation: V =q C, the voltage equations can be written as:
dt LdI C q C q
dt LdI C q C q
b a
=
−
=
−
3 2
2 1
so we have:
b a
I LC q q
I LC q q
&&
&
&
&&
&
&
=
−
=
−
3 2
2 1
i.e.
b a
I LC q q
I LC q q
&&
&
&
&&
&
&
=
−
=
−
3 2
2 1
By substitution of q&1 =−Ia, q&2 =Ia−Ib and q&3 =Ib into the above equations, we have:
b b
b a
a b
a a
I LC I I I
I LC I I I
&&
&&
=
−
−
= +
−
−
i.e.
0 2
0 2
= +
−
=
− +
b a b
b a a
I I I LC
I I I LC
&&
&&
By adding up and subtracting the above equations, we have:
0 ) (
3 ) (
0 )
(
=
− +
−
= + + +
b a b
a
b a b a
I I I
I LC
I I I I LC
&&
&&
&&
&&
Supposing the solutions to the above normal modes equations are given by:
t A I
Ia + b = cosω t B I
Ia − b = cosω so we have:
0 cos ) 3 (
0 cos ) (
2 2
= +
−
= +
−
t B
LC B
t A LC A
ω ω
ω ω
which are true for all t when
LC
2 = 1
ω and B=0
or
LC
2 = 3
ω and A=0
which show that the normal modes of oscillation are given by:
b
a I
I = at
LC
2 1
1 = ω and
b
a I
I =− at
LC
2 3
2 = ω
4.13
From the given equations, we have the relation between I1 and I2 given by:
1 2
2 I
L i Z
M I i
ω s
ω
= + so:
1 2
2 2
1 I
L i Z L M i MI i I L i E
s p
p ⎟⎟⎠
⎜⎜ ⎞
⎝
⎛ + +
=
−
= ω
ω ω ω
ω i.e.
s
p Z i L
L M I i
E
ω ω ω
+ +
=
2 2 2
1
which shows that E I1, the impedance of the whole system seen by the generator, is the sum of the primary impedance, iωLp, and a ‘reflected impedance’ from the
secondary circuit of ω2M2 Zs, where Zs =Z2+iωLs.
4.14
Problem 4.13 shows the impedance seen by the generator Z is given by:
s
p Z i L
L M i
Z ω
ω ω + +
=
2 2 2
Noting that M = LpLs and Lp Ls =n2p ns2, the impedance can be written as:
s p s
p s
s p p
L i Z
Z L i L
i Z
M M
Z L i L
i Z
M L
L Z
L Z i
ω ω ω
ω ω
ω ω
ω ω
ω
= + +
+
= − +
+
= −
2 2 2
2 2 2 2 2 2
2 2 2
2
so we have:
2 2 2 2
2
2 1 1 1 1
1
n Z n L Z i
L L L i Z
L i
L i Z Z
s p p s
p p p
s == + = +
= +
ω ω
ω ω
which shows the impedance Z is equivalent to the primary impedance iωLp connected in parallel with an impedance (np ns)2Z2.
4.15
Suppose a generator with the internal impedance of Z1 is connected with a load with an impedance of Z2 via an ideal transformer with a primary inductance of L and p the ratio of the number of primary and secondary transformer coil turns given by
s
p n
n , and the whole circuit oscillate at a frequency of ω . From the analysis in Problem 4.13, the impedance of the load is given by:
2 2 2
1 1
1
n Z n L i Z
s p p L
+
= ω
At the maximum output power: ZL =Z1, i.e.:
1 2 2
2
1 1
1 1
Z Z n L n i Z
s p p L
= +
= ω
which is the relation used for matching a load to a generator.
4.16
From the second equation, we have:
2 2
1 I
Z I Z
M
−
=
By substitution into the first equation, we have:
E I Z Z I
Z Z
M M
= +
− 1 2 2 2
i.e. 1
2 1
2 I
Z Z Z Z
I E
M
M ⎟⎟⎠
⎜⎜ ⎞
⎝
⎛ −
=
Noting that ZM =iωM and I2 has the maximum value when X1 = X2 =0, i.e.
1
1 R
Z = and Z2 = R2, we have:
1 2 1 1
2 1 1
2 1 1
2 1
2 2
2
I R R I E
M R M R I E
M R M R I E M i
R M R i
I E ≤ =
+
=
−
=
ω ω ω ω
ω ω
which shows I2 has the maximum value of 1
2
2 1 I
R R
E , when
M R M R
ω = ω1 2 , i.e.
2 1R R M = ω
4.17
By substitution of j=1 and n=3 into equation (4.15), we have:
2 0 2
0 2
0 2
1 (2 2)
2 1 2 4 2
cos 1
2ω π ω ω
ω ⎥ = −
⎦
⎢ ⎤
⎣
⎡ −
⎥⎦=
⎢⎣ ⎤
⎡ −
=
By substitution of j=2 and n=3 into equation (4.15), we have:
2 0 2
0 2
1 2
4 cos2 1
2ω π ω
ω ⎥⎦⎤=
⎢⎣⎡ −
=
By substitution of j =3 and n=3 into equation (4.15), we have:
2 0 2
0 2
0 2
1 (2 2)
2 1 2 4 2
cos3 1
2ω π ω ω
ω ⎥= +
⎦
⎢ ⎤
⎣
⎡ +
⎥⎦=
⎢⎣ ⎤
⎡ −
=
In equation (4.14), we have A0 = A4 =0 when n=3, and noting that ω02 =T ma, equation (4.14) gives:
when 1r= : 2 2 1 2 0
0 2
0 ⎟⎟⎠ − =
⎜⎜ ⎞
⎝
⎛ − +
−A A A
ω ω
i.e. 2 2 1 2 0
0
2 ⎟⎟ − =
⎠
⎜⎜ ⎞
⎝
⎛ − A A
ω
ω (4.17.1)
when 2r= : 2 2 2 3 0
0 2
1 ⎟⎟ − =
⎠
⎜⎜ ⎞
⎝
⎛ − +
−A A A
ω
ω (4.17.2)
when r=3: 2 2 3 4 0
0 2
2 ⎟⎟ − =
⎠
⎜⎜ ⎞
⎝
⎛ − +
−A A A
ω ω
i.e. 2 2 3 0
0 2
2 ⎟⎟⎠ =
⎜⎜ ⎞
⎝
⎛ − +
−A A
ω
ω (4.17.3)
Write the above equations in matrix format, we have:
0 2
1 0
1 2
1
0 1
2
3 2 1
2 0 2 2
0 2 2
0 2
⎟=
⎟⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
−
−
−
−
−
−
−
A A A ω ω ω
ω ω
ω
which has non zero solutions provided the determinant of the matrix is zero, i.e.:
0 ) 2
( 2 ) 2
( −ω2 ω02 3− −ω2 ω02 = The solutions to the above equations are given by:
2 0 2
1 (2 2)ω
ω = − ,` ω22 =2ω02, and ω12 =(2+ 2)ω02
4.18
By substitution of ω12 =(2− 2)ω02 into equation (4.17.1), we have:
0
2A1− A2 = i.e. A1:A2 =1: 2
By substitution of ω12 =(2− 2)ω02 into equation (4.17.3), we have:
0 2 3
2+ =
−A A i.e. A2:A3 = 2:1
Hence, when ω12 =(2− 2)ω02, the relative displacements are given by:
1 : 2 : 1 : : 2 3
1 A A =
A
By substitution of ω22 =2ω02 into equation (4.17.1), we have:
2 =0 A
By substitution of ω22 =2ω02 into equation (4.17.2), we have:
3 0
1+ =
−A A i.e. A1:A3 =1:−1
Hence, when ω22 =2ω02, the relative displacements are given by:
1 : 0 : 1 : : 2 3
1 A A = −
A
By substitution of ω22 =(2+ 2)ω02 into equation (4.17.1), we have:
0 2 1− 2 =
− A A i.e. A1:A2 =1:− 2
By substitution of ω12 =(2+ 2)ω02 into equation (4.17.3), we have:
0 2 3
2 − =
−A A i.e. A2 :A3 =− 2:1
Hence, when ω12 =(2+ 2)ω02, the relative displacements are given by:
1 : 2 : 1 : : 2 3
1 A A = −
A
The relative displacements of the three masses at different normal frequencies are shown below:
As we can see from the above figures that tighter coupling corresponds to higher frequency.
4.19
Suppose the displacement of the left mass m is x, and that of the central mass M is y , and that of the right mass m is z . The equations of motion are given by:
2 0
2 (2 2)ω
ω = − 02
2 2ω
ω = ω2 =(2+ 2)ω02
)
If the system has a normal frequency of ω, and the displacements of the three masses can by written as:
By substitution of the expressions of displacements into the above equations of motion, we have:
t
The matrix format of these equations is given by:
0
which has non zero solutions if and only if the determinant of the matrix is zero, i.e.:
0
The solutions to the above equation, i.e. the frequencies of the normal modes, are given by:
At the normal mode of ω2 =0, all the atoms are stationary, η1 =η2 =η3, i.e. all the masses has the same displacement;
At the normal mode of
m
= s
ω2 , η2 =0 and η1 =−η3 , i.e. the mass M is stationary, and the two masses m have the same amplitude but are “anti-phase” with respect to each other;
At the normal mode of
mM m M s( 2 )
2 = +
ω , η1:η2 :η3 =M :−2m:M , i.e. the two mass m have the same amplitude and are “in-phase” with respect to each other.
They are both “anti-phase” with respect to the mass M . The ratio of amplitude between the mass m and M is M 2m.
4.20
In understanding the motion of the masses it is more instructive to consider the range n
j
n 2≤ ≤ . For each value of the frequency ωj the amplitude of the rth mass is
sin 1
= +
n C rj
Ar π
where C is a constant. For j=n 2 adjacent masses have a π 2 phase difference, so the ratios: Ar−1:Ar :Ar+1=−1:0:1 , with the rth masses stationary and the amplitude Ar−1 anti-phase with respect to Ar+1, so that:
As n 2→n, Ar begins to move, the coupling between masses tightens and when j is close to n each mass is anti-phase with respect to its neighbour, the amplitude of each mass decreases until in the limit j= no motion is transmitted as the cut off n frequency ωj2 =4T ma is reached. The end points are fixed and this restricts the motion of the masses near the end points at all frequencies except the lowest.
4.21
By expansion of the expression of ω2j, we have:
→0 Ar
n j →
+1
Ar
−1
Ar
Ar
2 n j =
⎥⎦
⎢ ⎤
⎣
⎡ + −
+ + + −
⎟=
⎠
⎜ ⎞
⎝
⎛
− +
= L
! 6
) 1 (
! 4
) 1 (
! 2
) 1 (
2 cos 1
2 1 2 4 6
2 j n j n j n
ma T n
j ma
T
j
π π
π ω π
If n>>1 and j<< , n jπ n+1 has a very small value, so the high order terms of the above equation can be neglected, so the above equation become:
2 2 2
1
! 2
) 1 (
2 ⎟
⎠
⎜ ⎞
⎝
⎛
= +
⎥⎦
⎢ ⎤
⎣
⎡ +
= n
j ma
T n
j ma
T
j
π ω π
i.e.
ma T n
j
j = +π1 ω
which can be written as:
ρ
ω π T
l j
j = where, ρ =m a and l=(n+1)a
4.22
From the first equation, we have:
C q I q
L&&r−1= &r−1−&r
By substitution of q&r =Ir−1−Ir and q&r−1 =Ir−2−Ir−1 into the above equation, we have:
C I I I I
L&&r−1 = r−2−2 r−1+ r (4.22.1) If, in the normal mode, the currents oscillate at a frequency ω, we may write the displacements as:
t i r
r A e
I −2 = −2 ω , Ir−1 = Ar−1eiωt and Ir = Areiωt Using these values of I in equation (4.22.1) gives:
t r i r t r
i
r e
C A A e A
LA ω ω
ω = − +
− 2 −1 −2 2 −1
or
−Ar−2+(2−LCω2)Ar−1−Ar =0 (4.22.2) By comparison of equation (4.22.2) with equation (4.14) in text book, we may find the expression of Ir is the same as that of yr in the case of mass-loaded string, i.e.
t i t
i r
r e
n D rj e A
I ω π ω
sin 1
= +
=
Where D is constant, and the frequency ω is given by:
⎟⎠
⎜ ⎞
⎝
⎛
− +
= 1 1 cos 1
2
n j
j LC ω π
where, j=1,2,3...n
4.23
By substitution of y into 2
2
t y
∂
∂ , we have:
) ( 2 2
2 2 2
)
(ei teikx ei t kx t
t
y +
−
∂ =
= ∂
∂
∂ ω ω ω
By substitution of y into 2
2
x y
∂
∂ , we have:
) ( 2 2
2 2 2
)
(ei teikx k ei t kx x
x
y =− +
∂
= ∂
∂
∂ ω ω
If ω =ck, we have:
0 )
( 2 2 2 ( )
2 2 2 2 2
= +
−
∂ =
− ∂
∂
∂ i t+kx
e k x c
c y t
y ω ω
i.e.
2 2 2 2 2
x c y t
y
∂
= ∂
∂
∂