2214 Phys Homework #2 Solutions
September 9, 2011
1. Free oscillations
(a) For the initial conditions x(0) = A and v (0) = 0, we have the
following results:
i. Underdamped: x(t) = Aet/A cos(0 t)
ii. Overdamped: x(t) = Aet/A
iii. Critically damped: x(t) = A(1 + t/A )et/A
These are plotted in the gure.
1.0
0.5
0.0
2
4
6
8
10
Critically Damped
0.5
Underdamped
Overdamped
1.0
(b) For the underdamped case, since b2
4mk ,
b b2 4mk
=
2m
1
becomes
b i 4mk b2
.
=
2m
The imaginary part of is
4mk b2
2
2
Im() =
= 0 1/A .
2m
Since decreases when 0 A = 0, write 0 | | = 0 1 1/(0 A )2 .
Dividing by 0 , squaring both sides, and solving for 0 A gives
1
0 A =
2| |/0 (| |/0 )2
.
i. For | |/0 = 0.001, 0 A = 22.4.
ii. For | |/0 = 0.01, 0 A = 7.09.
iii. | |/0 = 0.1, 0 A = 2.29.
(c) For the underdamped case, the system undergoes many oscillations during the time it takes to relax to its equilibrium position.
For the critically damped and the overdamped cases, the system
doesnt undergo any oscillations.
i.
ii.
iii.
iv.
v.
vi.
vii.
Auto suspension: critically damped; slightly underdamped.
Door bell: underdamped.
Speaker cone: underdamped.
Pogo stick: underdamped.
Drum head: critical or underdamped.
Cymbal: underdamped.
Bungee cord + person: underdamped.
2. Driven oscillations
A(D ) =
2
0
F0 /m
+ i2D /A
2
D
(a) To nd the magnitude, use
A (D )A(D )
F0
1
=
2
m (0 2 )2 + 4 2 / 2
|A(D )| =
D
2
D
A
To nd the phase, rewrite A(D ) as
2
2 D i2D /A
1
F0
0
2
2
2
2
m 0 D + i2D /A 0 D i2D /A
2
2
F0 0 D i2D /A
=
2
2
2
2
m (0 D )2 + 4D /A
A(D ) =
Then
tan((D )) =
Im(A)
2D /A
.
=2
2
Re(A)
0 D
(b) We can maximize |A(D )| by minimizing the denominator:
d
2
2
2
2
(0 D )2 + 4D /A = 0
dD
2
2
2
2(0 D )(2D ) + 8D /A = 0 ,
so D =
2
2
0 2/A . Since Q = 0 A /2, we get D = 0 1 1/2Q2 .
(c) Plug D from part (b) into the complex amplitude to get:
|A|max =
F0
m
1
=
F0
m
=
F0
2
m 0 1 1/4Q2
2
2
2
2
(0 0 (1 1/2Q2 ))2 + 40 (1 1/2Q2 )/A
1
2
4
(0 /2Q2 )2 + 0 (1 1/2Q2 )/Q2
Q
2
Using m0 = k and A0 = F0 /k gives Amax =
A0 Q
11/4Q2
A0 Q for
large Q.
(d) For large Q, D
0 = k /m. To nd how changes in mass
change the frequency, dierentiate with respect to m:
d
d
1
1
=k
(m1/2 ) = k ( )m3/2 =
dm
dm
2
2
Then | |/
m/2m. But | |/
4
2/Q = 2/10 = 0.0002.
3
k1
1
=
.
mm
2m
1/Q, so m/m
(e) When D = 0 ,
F0 /m
F0
F0 m
A = |A(0 )| =
=
=.
20 /A
b0
bk
Dierentiating A with respect to m gives
dA
F0 1
=
dm
2b k m
A
m
=
.
A
2m
and
So, for a change in mass of 1 part in 106 , A/A = 5 107 .
(f) Suppose Q = 1/ 2. Then max = 0 1 1/2Q2 = 0. This
means
F0 Q
F0
Amax =
= A0 .
2
2
m0
m0 1 1/4Q2
So, Amax = A0 , which means that the resonance peak equals the
amplitude at D = 0 and the peak goes away.
3. Free and driven oscillations
For t = 3A , the amplitude decreases by e3 = 0.05, which is close to
zero.
(a) Underdamped, since you are told that the oscillations die away.
(b) Given f0 = 4Hz , we have 0 = 25.13 rad/s. Since 3A = 10s, then
A = 10/3 s. So, Q = 0 A /2 = 41.9, and D = 0 1 1/2Q2 =
25.13 rad/s 0 .
(c) A = 10/3 s, f0 = 4 Hz 0 = 8 rad/s.
A(d ) =
F0 /m
2
2
2
2
(0 D )2 + 4D /A
,
so Amax = (F0 /m)/(2D /A ). Then we set A = 0.7 Amax and
solve for D .
0.7
F0 /m
=
2D /A
F0 /m
2
2
2
2
(0 D )2 + 4D /A
.
Squaring both sides and inverting 4D gives
2
2
2
2
2
2.041 /A = (0 D )2 + 4d /A .
4
2
Rearranging terms gives a quadratic equation for D whose solu2
2
2
tions are D = (6.476 10 , 6.161 10 ). Then D = (24.82, 25.45)
and D = 0.63 rad/s.
(d) At low frequencies, |A|max /A0 = Q = 41.9. At high frequencies,
|A|max /|A(D ) .
(e) Removing the cashews decreases the mass of the oscillator. Then
0 = k /m increases. The damping constant doesnt change,
so Q = 0 A /2 = k m/b decreases. Since 0 /0 = 1/Q,
0 = 0 /Q = b/m, so 0 increases.
4. Power in driven oscillators
If the driving force is F0 cos(t) and the position is given by x(t) =
A cos(t + ), then v (t) = A sin(t + ) and the instantaneous
power delivered to the oscillator by the driving force is
PD (t) = F (t) v (t) .
Some useful integrals for this problem are:
2
2
sin(x) cos(x)dx = 0
and
0
0
2
sin2 (x)dx =
cos2 (x)dx = .
0
(a) The average power delivered by the driving force in one cycle (the
period is 2/ ) is:
2/
F0 cos(t)(A) sin(t + )dt
2 0
AF0 2
=
cos y sin(y + )dy
2 0
AF0 2
=
[cos y sin y cos + cos2 y sin ]dy
2 0
AF0
=
sin .
2
avg
PD
=
(b) The instantaneous power dissipated by the damping force is Pdamp (t) =
bv 2 . The average power delivered per cycle is:
2/
(A )2 sin2 (t + )dt
2 0
1
= b 2 A2
2
avg
Pdamp = b
5
(c) The frequency at which the input power is maximized is obtained
by maximizing the answer in part (b):
dPdamp
d
2
2
d 2
d 2 ( 2 0 )2 + b2 2 /m2
2
2
[( 2 0 )2 + b2 2 /m2 ] 2 [2( 2 0 ) + b2 /m2 ]
2
[( 2 0 )2 + b2 2 /m2 ]2
=0
Setting the numerator equal to 0 gives
2
2
( 2 0 )2 2 2 ( 2 0 ) = 0
4
2
2
4 + 0 2 2 0 2 4 + 2 2 0 = 0
avg
avg
Then Pdamp , and thus PD , is a maximum when = 0 .
The phase ( ) is given by
tan =
2
b
=
.
2
2
2)
A (0
m(0 2 )
Then ( = 0 ) = |i/2 and v (t) = A sin(t /2) =
A cos t. So, when = 0 , the driving force and the velocity
are in phase.
5. Energy in driven oscillations
For a system udergoing forced oscillations at an angular frequency D ,
x(t) = |A(D )| cos(D t + ) ,
where
|A(D )| =
F0
m
1
2
2
2
2
(0 D )2 + 4D /A
=
2
A0 0
2
2
2
2
(0 D )2 + 4D /A
;
In the last step we used F0 /m = (F0 /m) (k/k ) = (F0 /k ) (k/m) =
2
A0 0 . Also,
v (t) = D |A(D )| sin(D t + ) .
6
(a) The kinetic energy is K (t) = mv (t)2 /2, so
1
2
K (t) =
mD |A(D )|2 sin2 (D t + )
2
4
0
1
2
mD A2 2
sin2 (D t + )
=
0
22
2
2
2
(0 D ) + 4D /A
(b) The potential energy is U (t) = k (x x0 )2 /2, so
2
1
A0 0
U=
A0
k2
2
2 0 D + 2iD /a
2
2
1
0
2
=
m0 A2 2
1
0
2
2
0 D + 2iD /A
1
2 2iD /A
2
=
m0 A2 2 D 2
0
2
0 D + 2iD /A
2
2
Then
4
1
D + (2D /A )2
22
|U| = m0 A0
2
2
2
(0 D )2 + (2D /A )2
and U (t) = |U| cos2 (D t + ).
,
(c)
22
2
Kmax
0 D
0
1
=4
,
=2
=2
2
2
2
Umax
D + 4D /a
D + 4/A
x + 1/Q2
where x = D /0 and Q = 0 A /2.
(d) Kmax = Umax when the ratio in part (c) is equal to 1. Then
2
2
2
0 = D + 4/A , or
D =
2
2
0 4/A = 0 1 1/Q2 .
Then the total energy, which is equal to the maximum kinetic
energy, is:
4
0
1
2
2
m(0 4/A )A2 2
0
2
2
2
2
2
2
[0 (0 4/A )2 + 4(0 4/A )/A
1
4 ( 2 4/ 2 )
=
mA2 0 0 2 2 A
0
2
40 /A
1
2
=
mA2 0 Q2 (1 1/Q2 )
0
2
1
2
=
m0 A2 (Q2 1)
0
2
E = Kmax =
7
(e) For low driving frequencies, D
Q2
1, so Umax
Kmax .
22
0 and Kmax /Umax 0 A /2 =
(f) For high driving frequencies, D and Kmax /Umax
Umax
Kmax .
8
1, so