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Chapter 2
OSCILLATIONS
Topic
Oscillatory Motion
Simulations
Bungee jumps and bridge
swings
m-script
bridge_swing.m
We our surrounded by a world of oscillations. Our heart pumps to a regular rhythm; the
air vibrates when we talk; we can see because light waves are oscillating electric fields
they stimulate the receptors on the retina. We enjoy listing to music; the strings vibrate on
a guitar when plucked; the air vibrates when you play a trumpet and the membrane of a
drum vibrates when struck.
To start the study of oscillations we will consider an object of mass m attached to a spring
with a stiffness k . The object can be displaced from its equilibrium and set to oscillate
freely or it can be forced to oscillate by some mechanical means. As the object oscillates
it can dissipate energy to the surroundings. This dissipation of energy is called damping.
The equation of motion for our object confined to oscillate along the x-axis is derived
from Newton’s Second Law. We will assume that the elastic limit of the spring is never
exceeded. The forces acting of the object are the elastic restoring Fe forces, the damping
force Fb responsible for the energy dissipation and the driving force Fd which imparts
energy to the oscillating object. We will assume the elastic restoring force is given by
Hok’s law and a velocity dependence for that damping force and consider the driving
force to be sinusoidal (harmonic) and given by a step function. The equation of motion is
given by
(1)
m a(t )  k x(t )  bv(t )  Fd (t )
where x(t) is the displacement from the equilibrium position (x = 0).
The numerical procedure adopted, is based on a half-step method and the smaller the time
interval t used the better the approximations. From the definitions of velocity and
acceleration, at any time, t we have
(2)
v (t ) 
x (t  t / 2)  x (t  t / 2)
t
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(3)
a (t ) 
x (t  t )  2 x (t )  x (t  t )
t 2
Therefore, the displacement of the object at time (t + t) can be calculated directly from
the displacement at the two previous time steps, t and (t - t)
(4)
x(t  t )  c1 x(t )  c2 x(t  t )  c3 Fd
where the constants c1, c2 and c3 are given by
b t
c0  1 
2m
2  k t 2 / m
c1 
c0
b t /  2m   1
t 2 / m
c2 
c3 
c0
c0
Simple Harmonic Motion
For the case of simple harmonic motion, the damping and the driving force are zero and
for damped harmonic motion only the driving force is zero.
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Ther
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Simulation 2.1
Bungee jumps and bridge swings
Inspect and run the m-script bridge_swing.m so that you are familiar with what the
program and the code does. The m-script calls the function eq_quadratic.m.
A bungee jump is when a person plummets off some high structure such as a tower,
bridge, crane or hot air balloon. A version of the bungee jump is the bridge swing. A
bungee cord is connected to one side of a bridge and the other end to the jumper in an
abseiling harness on the other side of the bridge. Bungee cords are soft and springy and
may stretch to more than three times there natural length. Most jumps occur without any
mishap, however, there have been many serious injuries and deaths, and in some
countries bungee jumping is illegal. Accidents can be traced to human error such as
improper attachment of the body harness to the jumper and others to the poor
understanding of the physics where there has been a mismatch between the mass of
jumper, height of jump and bungee cord selected for the jump.
Simple models based upon Newton’s Second Law (equation of motion) can be used to
give an insight into the physics of bungee jumping and to give estimates of jump drop,
maximum speed, acceleration, forces and energies during a jump. A two dimensional
model is used for the simulations of bungee jumps and bridge swings. The coordinate
system is one in which the +y-axis is vertical up and the +x-axis is horizontal to the right.
The origin (0,0) is the point of attachment of the bungee cord to some structure. The
jumper is assumed to be a point particle of mass m and during the jump, the forces acting
on the jumper are the gravitational force FG , the elastic restoring force due to the
extension of the bungee cord, FE and a dissipative force, FD due to friction and drag
forces acting on and within the cord and on the jumper. The equation of motion for the
jumper is
(1)
F  F
(2)
r  L  e  x2  y2 .
FE  FD  ma
where a is the acceleration of the jumper and the position of the jumper has coordinates
(x, y). The natural length of the bungee cord is L, the extension of the cord, e and
extended length of the cord, r
G
The bungee cord acts like a weak spring with a stiffness that decreases with extension
(Menz, 1993). A two-piece linear model is used to describe the stiffness of the bungee
cord; the stiffness is k1 when the extension is less than e1 and k2 for extension greater than
e1 where k1 > k2. The elastic restoring force when the cord is stretched is given by
(3a)
0  e  e1
(3b)
e  e1
FE  k1 e
FE  k2 e   k1  k2  e1
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(k1  constant)
(k1, k2  constant) .
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The dissipative force is assumed to depend upon the both the velocity and the square of
the velocity of the jumper. The x and y components of the forces acting on the jumper are
given in equation (3)
(4a)
FGx  0
(4b)
FEx   FE
(4c)
FDx   D1vx  D2vx v
FGy  m g
x
r
FEy   FE
y
r
FDy   D1v y  D2v y v
where D1 and D2 are constants for the dissipative force and the elastic restoring force, FE
is given by equation (3).
The x and y components for the acceleration of the jumper are then
(4a)
ax 
 FGx  FEx  FDx 
m
and
ay 
F
Gy
 FEy  FDy 
m
.
A simple numerical method is used to calculate the velocity and displacement from the
acceleration at successive time steps given a set of initial conditions. The method outlined
is not the most accurate method but one in which is easy to implement. If the values of
x(t), y(t), vx(t), vy(t), ax(t) and ay(t) are known at time t then the values at time t + t are
(5a)
v x (t  t )  v x (t )  a x (t ) t
x(t  t )  x (t )  v x (t ) t  0.5 a x (t )t 2
(5b)
v y (t  t )  v y (t )  a y (t ) t
y(t  t )  y(t )  v y (t ) t  0.5 a y (t )t 2 .
Once the acceleration, velocity and displacement are known, it is a simple matter to
calculate forces and energy. The kinetic energy of the jumper, K, the elastic potential
energy, UE, the gravitational potential energy, UG and the total mechanical energy, E are
given in equation (6). The zero for the gravitational potential energy is chosen to be at y =
0.
K  12 m v 2
(6a)
(6b)
(6c)
(6d)
UG   m g  L  e 
0  e  e1 U E  12 k1 e 2
e  e1 U E  12 k2 e 2   k1  k2  e1 e  12  k1  k2  e12
E  K  U E  UG .
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If the dissipative force is taken as zero, the total mechanical energy is conserved. From
the principle of conservation of energy in this situation, we can predict the maximum free
fall velocity, vff and the maximum distance the jumper falls vertically (jump drop), hdrop.
the total energy at three instances for a bungee jump are
(7a)
start of jump
E1  K1  U E1  UG1  0  0  0
(7b)
end of free fall
E2  K 2  U E 2  U G 2  12 m vffmax 2  0  m g L
(7c)
bottom of jump
E3  K3  U E 3  UG3  0  U E 3  m g  L  emax  .
From equations (7a) and (7b), the maximum free fall velocity is
(8)
v ff  2 g L
Equating equations (7a) and (7c) gives a quadratic equation for the maximum extension,
emax and which can be solved to give the maximum extension of the bungee cord and the
drop height, hdrop
(9a)
2
1
emax 2      k1  k2  e1  mg  emax      k1  k2  e12  2 m g L   0
 k2 
 k2 
(9b)
hdrop  L  emax .
One has to be wary in using numerical method because you are not always confident that
the results are accurate. The approximation method used in this simulations only gives
reasonable results when the time increment is very small so that during any time step the
acceleration does not change very much. When the time step is too large, the total
mechanical energy often increases rather than stay constant. Where possible you should
compare your numerical results with values calculated using analytical methods. If there
is disagreement, then the time step used in the numerical procedure can be decreased or it
may be necessary to use a more accurate numerical method. Table 1 gives a set of
possible ranges for the parameters describing a jump.
Variable
mass of jumper
natural length of bungee cord
stiffness of bungee cord:
stiff cord
stiffness of bungee cord:
medium cord
stiffness of bungee cord:
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Symbol
m
L
k1
k2
k1
k2
k1
Value
50 to 150
6 to 20
~ 250
~ 150
~ 200
~ 110
~ 160
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Unit
kg
m
N.m-1
N.m-1
N.m-1
6
soft cord
stiffness of bungee cord
max tensile strength of bungee cord
elongation factor
dissipation factor
dissipation factor
bridge swing: starting x coordinate
Table 1: Jump parameters
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k2
e1
Smax
e/L
D1
D2
x(0)
~ 80
4 to 6
~ 2.0104
2 to 4
0 to 10
0 to 10
0 to - 10
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m
N
N.m-1.s-1
N.m-2.s-2
m
7
Sample results:
bridge_swing.m
Bungee Jump
Input parameters
Initial x coordinate of jumper, xs = 0.00 m
Initial y coordinate of jumper, ys = 0.00 m
Mass of jumper, m = 100.00 kg
Natural length of bungee cord, L = 9.00 m
Dissipative force constant, D_1 = 0.00 N.m^-1,s^-1
Dissipative force constant, D_2 = 0.00 N.m^-2,s^-2
stiffness, k_1 = 200.00 N.m^1
stiffness, k_2 = 200.00 N.m^1
stiffness: two piece linear transition, e_1 = 0.00 m
Analytical & numerical predictions
A: max free fall velocity, v_ff = 13.28 m/s
N: max free fall velocity, v_ff = 13.28 m/s
A: drop height, h_drop = 24.49 m
N: drop height, h_drop = 24.51 m
Output parameters
max velcoity, v_max = 15.01 m.s-1
max velcoity, v_max = 54.03 km.h-1
max acceleration, a_max / g = 2.17
max force on jumper, F_max = 2122 N
max elastic restoring force by bungee cord, F_E_max = 3102 N
1
position x (m)
0
0
-0.5
-1
-10
-20
5
10
15
time t (s)
20
25
0
5
10
15
time t (s)
20
25
0
-10
-20
-30
-10
-5
0
x position (m)
5
10
15
2.5
5000
0
-5000
4000
2000
0
5
0
10
5
0
15
time t (s)
10
5
10
20
15
time t (s)
20
15
time t (s)
x 10
4
2
K
U
1.5
U
1
E
G
25
25
20
25
energy (J)
0
Fynet (N)
4
2
0
Fe (N)
a/g
-15
E
0.5
0
-0.5
-1
-1.5
4000
Fe (N)
0
10
-15
position y (m)
y position (m)
-5
0.5
2000
-2
0
-2.5
0
2
4
6
8
10
extension, e (m)
12
14
16
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0
5
10
15
time t (s)
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Bridge Swing
Input parameters
Initial x coordinate of jumper, xs = -5.00 m
Initial y coordinate of jumper, ys = 0.00 m
Mass of jumper, m = 100.00 kg
Natural length of bungee cord, L = 9.00 m
Dissipative force constant, D_1 = 4.00 N.m^-1,s^-1
Dissipative force constant, D_2 = 4.00 N.m^-2,s^-2
stiffness, k_1 = 200.00 N.m^1
stiffness, k_2 = 110.00 N.m^1
stiffness: two piece linear transition, e_1 = 5.00 m
Analytical & numerical predictions
A: max free fall velocity, v_ff = 13.28 m/s
N: max free fall velocity, v_ff = 10.34 m/s
A: drop height, h_drop = 28.10 m
N: drop height, h_drop = 21.29 m
Output parameters
max velcoity, v_max = 11.02 m.s-1
max velcoity, v_max = 39.69 km.h-1
max acceleration, a_max / g = 1.00
max force on jumper, F_max = 980 N
max elastic restoring force by bungee cord, F_E_max = 1805 N
5
position x (m)
0
-2
-4
-8
-5
-12
-16
-18
-20
0
x position (m)
5
5
0
10
5
0
15
time t (s)
10
5
10
20
15
time t (s)
20
25
0
5
10
15
time t (s)
20
25
x 10
4
K
U
1
25
G
U
E
0.5
15
time t (s)
20
15
time t (s)
20
25
energy (J)
a/g
Fynet (N)
Fe (N)
2000
1000
0
10
-20
1.5
1000
0
-1000
5
-10
-30
-5
0
0
0
-14
1
0.5
0
0
-10
position y (m)
y position (m)
-6
E
0
-0.5
-1
25
-1.5
Fe (N)
2000
-2
1000
0
0
2
4
6
8
extension, e (m)
10
12
14
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-2.5
0
5
10
15
time t (s)
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20
25
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Investigations and Questions
Simulate a bungee jump (start: xs = 0, ys = 0) with a zero dissipation force.
1.1 Vary the number of time steps, nt. What is the minimum value for nt so that the
total mechanical energy, E is conserved. How well do the analytical and numerical
predictions agree?
1.2 Explain the shapes of the position, force and energy graphs. Comment on when
values of position, force, energy are a maximum, minimum or zero.
1.3 For a typical bungee jump, is the acceleration experience by the jumper excessive?
(A person can blackout when their acceleration greater than about 5g’s).
1.4 Is it likely that a bungee cord will snap? (The maximum tensile strength of a bungee
cord is ~2.0104 N.)
1.5 How does the vertical distance fallen by the jumper (jump drop), hdrop, depend upon
the mass of the jumper, m? Repeat the simulation with increasing mass and tabulate
the jump drop.
1.6 What are the differences in jump drops for bungee cords of varying stiffness.
Consider stiff, medium and soft bungee cords.
1.7 Simulate a bungee jump with a non-zero dissipative force. How does this change
the motion of the jumper? Adjust the value of the constants D1 and D2 to simulate
as close as possible a real jump.
1.8 What is the mostly likely reason a person could suffer a serve accident assuming
there where no attachment problems associated with the harness and cord?
1.9 If the stiffness of the bungee cord obeys Hooke’s law in extension (e1 = 0 and k1 =
k2), how does it affect the jump drop. Is this significant?
Simulate a bridge swing with zero and non-zero dissipative forces. Check that the time
step is small enough so that the total energy is conserved and that the numerical and
analytical values agree.
1.10 Explain the shapes of the position, force and energy graphs. Comment on when
values of position, force, energy are a maximum, minimum or zero. The figure for
the trajectory shows the position of the jump at fixed time intervals like in a strobe
photograph, hence comment on the velocity of the jumper during the swing.
1.11 You can make your own bungee cord by using a length of latex tubing or by
stringing together many elastic bands. Use you home-made bungee cord with an
object attached to one end to carry out a bridge swing. How does the trajectory of
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your bride swing compare with the computer model? You can change the
parameters in the simulation to match those in your bungee swing using your home
made cord.
1.12 Design a jump so that the maximum velocity of the jumper exceeds 80 km.h-1.
What is the corresponding jump drop?
References
Menz, P.G.,, The Physics of Bungee Jumping, The Physics Teachers, Vol. 31, No. 8,
November 1993.
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Miscellaneous
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