Physics 102
Waves
Lecture 2
More Oscillations
Moza M. Al-Rabban
Professor of Physics
mmr@qu.edu.qa
The Dynamics of SHM
x(t )  A cos t  0 
d2
ax (t )  2  A cos t  0  
dt
  2 A cos t  0 
  x(t )
2
The acceleration is proportional
to the negative of the displacement
at any time.
2
The Equation of Motion in SHM
( Fsp ) x  k x
(Fnet ) x  ( Fsp ) x  kx  max
k
ax   x
m
d 2x k
 x0
2
dt
m
This is the equation of motion for the system. It is a homogeneous
linear 2nd order differential equation.
3
Solving the Equation of Motion
d 2x k
 x0
2
dt
m
Q: How do you solve this differential equation?
A: Our method - guess the solution, and see if it works.
x  A cos t  0 
 2 A cos t  0  
d
d
x   A cos t  0     A sin t  0 
dt
dt
d2
d
2
x



A
sin

t





A cos t  0 




0
2


dt
dt
2 
k
A cos t  0   0
m
k
k
or  
m
m
It works!
4
Example:
Analyzing an Oscillator
At t=0, a 500 g block oscillating on a
spring is observed to be moving to the right
at x=15 cm. At t=0.30 s, it reaches its
maximum displacement of 25 cm.
x  A cos
(a) Draw a graph of the motion for one cycle.
(b) At what time in the first cycle is x=20 cm?
The motion is SHM. The position equ. Of the block is
x (t )  A cost   
x0  15 cm, A  25 cm
x  A cos
0  cos1  x0 / A  cos1  0.60  0.927 rad
The object is initially moving to the right, which tells us
that the phase constant must be between - and 0 rad.
Thus   0.927rad
5
Example:
Analyzing an Oscillator
(a) Draw a graph of the motion for one cycle.
(b) At what time in the first cycle is x=20 cm?
The motion is SHM. The position equ. Of the block is
x (t )  A cost   
x max  A  A cos(wt   )
The block reaches its maximum displacement at time
t = 0.30 s. At that instant of time
At t  0.30 s, cos t  0   1
Therefore,   0 / t  (0.927 rad)/0.30 s  3.09 rad/s
T  2 /   2.03 s
x (t )  (25cm ) cos(3.09t  0.927)
t    cos 1 ( x / A)  0  / 
 0.0917 s and 0.508 s
6
Vertical Oscillations
Does the additional force of gravity change the
motion?
Now consider a mass m hanging from a spring
with constant k. When the mass is attached, the
spring stretches by L.
Finding L is a static –equilibrium problem in
which the upward spring force balances the
downward weight force of the block. So,
F 
sp y
 ky  kL
 L is a distance and it is a +ve number.
y is the displacement.
In this configuration, the equilibrium position is
given by:
At equilibrium, (Fnet ) y  (Fsp ) y  wy  k L  mg  0
L  mg / k  equilibrium position
7
Let the block oscillate around this equilibrium position.
The origin of the y-axis at the block’s equilibrium position
If the block moves upward, the spring gets
shorter compared to its equilibrium length,
but the spring is still stretched compared
to its unstretched length.
When the block is in position y, the spring
is stretched by an amount L  y
m
And hence exerts an upward spring force
Fsp  k L  y 
The net force on the block at this point is
(Fnet ) y  ( Fsp ) y  wy  k (L  y)  mg  (kL  mg )  ky
( Fnet ) y  ky  y(t )  A cos(t  0 )
  k/m
Because, the motions are the same,
everything we have learned about
horizontal oscillation is equally valid for vertical oscillations.
8
Example 7: A Vertical Oscillator
A 200 g block hangs from a spring
with constant k = 10 N/m. The block is
pulled down to a point where the spring
is 30 cm longer than its unstretched
length, then released.
Where is the block and what is its
velocity 3.0 s later?
Although the spring begins by being stretched 30 cm, this is not the amplitude
of the oscillation. Oscillations occur around the equilibrium position, so we
have to begin by finding the equilibrium point where the block hangs
motionless on the spring.
L  mg / k  (0.20 kg)(9.8 m/s2 )/(10 N/m)  0.196 m
Stretching the spring 30 cm pulls the block 10.4 cm below the
equilibrium point,
A  ys  L  (0.300 m)  (0.196 m)  0.104 m
so A=10.4 cm.
9
Example 7: A Vertical Oscillator
y(t )  10.4cm coswt   
  k / m  (10 N/m) /(0.20 kg)  7.07 rad/s
The initial condition
y  A  A cos   
At t= 3.0 s the block’s
position is:
y  (10.4cm ) cos((7.07rad / s)(3.0 s)   rad )  7.4cm
The block is 7.4 cm above the equilibrium position, or 12.2 cm below the
original end of the spring.
Its velocity at this instant is
v x  wA sin(wt   )  52cm / s
10
The Pendulum
The pendulum’s position can be described by the
arc of length s, which is zero when the pendulum
hangs straight down.
Because angles are measured ccw, s and  are
+ve when pendulum is to the right of center,
-ve when it is to the left.
11
The Pendulum
Two forces are acting on the mass:
The string tension and the weight.
( Fnet )t   Ft  wt
 mg sin   mat
d 2s
Using at  2
dt
for acceleration “ around” the circle,
d 2s
  g sin 
2
dt
Where the angle  is related to the arc length by =s/L,
so
d 2s
s

g
sin
0
2
dt
L
The equation of
motion for an
oscillating pendulum.
The sine function makes this equation more complicated than the
oscillating spring’s ones.
This is not a homogeneous linear 2nd order differential equation
describing SHM.
12
The Small Angle Approximation
s  r
The height of the
triangle is h=r sin 
Suppose that the angle
 is “small”.
sin   
( in radians )
The pendulum equation is:
d 2s
s

g
sin
0
2
dt
L
d 2s g
 s0
2
dt
L
This is a homogeneous linear 2nd order differential
equation describing SHM, but only for small swings (s<<L).
13
The Small Angle Approximation
d 2s g
 s0
2
dt
L
This is exactly the same
equation as the equation
for mass oscillating on a
spring.
s (t )  A cos(wt   )
or
The angular frequency
Is determined by the length
of the string.
d 2x k
 x0
2
dt
m
 (t )   max cos(wt   )
g
w  2f 
L
14
Example 8: A Pendulum Clock
What length L must a pendulum have to
give an oscillation period T of exactly 1 s?
T  2
L
(independent of m)
g
2
2
T 
2 1 s 
L  g
  (9.80 m/s ) 
  0.248 m
 2 
 2 
15
Example 9: The
Maximum Angle of a Pendulum
A 300 g mass on a 30 cm long string oscillates as
a pendulum. It has a speed of 0.25 m/s as it passes
through the lowest point.
What maximum angle does the pendulum reach?

g
(9.80 m/s2 )

 5.72 rad/s
L
(0.30 m)
The speed of the lowest point is:
vmax   A
A  vmax /   (0.25 m/s)/(5.72 rad/s)  0.0437 m
 max 
A (0.0437 m)

 0.145 rad  8.3
L
(0.30 m)
16
Conditions for SHM
we have solved all simple harmonic motion problems once we have
solved the problem of horizontal spring.
The restoring force of a spring, Fsp = kx, is directly proportional
to the displacement x from equilibrium.
The pendulum’s restoring force, in the small angle approximation, is
directly proportional to the displacement s.
A restoring force that is directly proportional to the displacement
from equilibrium is called a linear restoring force.
For any linear restoring force, the equation of motion is identical
to spring equation ( other than perhaps using different symbols).
Any system with a linear restoring force will
undergo simple harmonic motion around the
equilibrium position.
17
Everything that we learn about an oscillating spring can be applied
to the oscillations of any other linear restoring force, ranging from
the vibration of airplane wings to the motion of electrons in electric
circuits.
Moreover, even when the restoring force is nonlinear (e.g., the
pendulum), for sufficiently small displacements it can be treated
as a linear system (small angle approximation).
Therefore, we often treat atoms, molecules, and nuclei as
small harmonic oscillators, even when the restoring forces
involved may be very different from Hooke’s Law forces.
18
Identifying and Analyzing SHM
1. If the net force acting on a particle is a linear restoring force, the motion
will be simple harmonic motion (SHM) around the equilibrium position.
2. The position x as a function of time is x(t) = A cos (t + 0).
The velocity v as a function of time is v(t) =  A sin (t + 0).
The maximum speed is vmax = A .
The equations are given here in terms of x, but they can be written in
terms of any other appropriate variables (y, , etc.)
3. The amplitude A and the phase constant 0 are determined by the initial
conditions, through x0 = A cos 0 and v0x = A sin 0.
4. The angular frequency  and the period T depend on the physics of the
particular situation but are independent of A and 0.
5. Mechanical energy is conserved. Thus, ½mvx2 + ½kx2 = ½kA2 = ½mvmax2.
Energy conservation provides a relationship between position and velocity
that is independent of time.
19
Clicker Question 1
One person swings on a swing and finds that the period is 3.0 s.
A second person of equal mass joins him on the same swing.
With two people swinging, the period is:
a. 6.0 s;
b. 3.0 s;
c. 1.5 s;
d. Between 3.0 s and 6.0 s;
e. Can’t tell without knowing the length.
20
Damped Oscillations
Drag force:
b: the damping
constant
D  bv
(Fnet )x  (Fsp ) x  Dx  kx  bvx  max
d 2 x b dx k

 x0 
2
dt
m dt m

2
k
b
1
 2  02  2
m 4m
4
x(t )  A e

bt
2m
cos(t  0 )
Ampl. Damping
factor
Sinusoidal
variation
k
m
0 
and  
m
b
(damped oscillator)
Note that
damping reduces
the oscillation
frequency.
21
Energy in Damped Systems
x(t )  A et / 2 cos(t  0 );  
m
b
2
E (t )  12 kxmax
 12 k ( Aet / 2 )2


1
2

kA2 et /  E0et /
The oscillator’s mechanical
energy decays exponentially
with time constant .
e  2.71828...
e 1  0.3678...  36.8%
e 2  0.1353...  13.5%
e 3  0.04978...  4.98%
22
Example: A Damped Pendulum
A 500 g mass on a 50 cm long string oscillates as a
pendulum. The amplitude of the pendulum is observed
to decay to ½ of its initial value after 35 s.
(a) What is the time constant  of the damped oscillator?
(b) At what time t1/2 will the energy of the system have
decayed to ½ of its initial value?
At t  0, xmax  A, and at t  35 s, xmax  A / 2.
Aet / 2  A / 2  t / 2  ln 2
  t / 2ln 2  (35 s) / 2ln 2  25.2 s
E0et1/ 2 /  E0 / 2  t1/ 2 /   ln 2
t1/ 2   ln 2  (25.2 s)ln 2  17.5 s
23
Clicker Question 2
Which of these decays has the shortest time constant?
24
Driven Oscillations
and Resonance
Now, suppose we drive a damped mechanical
oscillator with an external force F(t) = Fd cos(dt).
Then the equation of motion is:
Fd
d 2 x b dx k


x

cos d t
2
dt
m dt m
m
The system will show the property of
resonance. The oscillation amplitude will
depend on the driving frequency d, and will
have its maximum value when: d  0  k / m
i.e., when the system is driven at its resonant
frequency 0.
The width of the resonance curves depends on
the amount of damping. Wider curves with smaller
resonance curves correspond to more damping and
larger values of b.
25
Chapter 14 Summary (1)
26
Chapter 14 Summary (2)
27
Chapter 14 Summary (3)
28
End of Lecture 2
 Before the next lecture, read
Knight, Chapters 20.1 through 20.3.
Homework #1
1.
In taking your pulse, you count 75 heartbeats in 1 min. What are the period
and frequency of your heart’s oscillations?
5.
What are the (a) amplitude,
(b) frequency, and (c) phase constant of the
oscillation shown in the Figure.
14.
The position of a 50 g oscillating mass is given by
x (t )  (2.0cm ) cos(10t   / 4) , where t is in s. Determine:
a. The amplitude.
b. The period.
c. The spring constant.
d. The phase constant.
e. The initial conditions.
f. The maximum speed.
g. The total energy.
h. The velocity at t = 0.4 s.
30
18. A spring is hanging from the ceiling. Attaching a 500 g physics book to the spring causes it
to stretch 20 cm in order to come to equilibrium.
a. What is the spring constant?
b. From equilibrium, the book is pulled down 10 cm and released. What is the period of
oscillation?
c. What is the book’s maximum speed? At what position or positions does it have this
speed?
21. Make a table with 4 columns and 8 rows. In row 1, label the columns  (°),  (rad), sin ,
and cos  . In the left column, starting in row 2, write 0, 2, 4, 6, 8, 10, and 12.
a. Convert each of these angles, in degrees, to radians. Put the results in column 2. Show
4 decimal places.
b. Calculate the sines and cosines. Put the results, showing 4 decimal places, in columns
3 and 4.
c. What is the first angle for which  and sin  differ by more than 0.0010?
d. What is the first angle for which cos  is less than 0.9900?
e. Over what range of angles does the small-angle approximation appear to be valid?
31
29. A 205 g air-track glider is attached to a spring with spring constant 4.0 N/m. The damping
constant due to air resistance is 0.015 kg/s. The glider is pulled out 20cm from equilibrium and
released. How many oscillations will it make during the time in which the amplitude decays to
e 1 of its initial value?
32