Download Chapter 15

Chapter 15
Oscillations
Periodic motion
• Periodic (harmonic) motion – self-repeating motion
• Oscillation – periodic motion in certain direction
• Period (T) – a time duration of one oscillation
• Frequency (f) – the number of oscillations per unit
time, SI unit of frequency 1/s = Hz (Hertz)
1
f 
T
Heinrich Hertz
(1857-1894)
Simple harmonic motion
• Simple harmonic motion – motion that repeats itself
and the displacement is a sinusoidal function of time
x(t )  xm cos(t   )
Amplitude
• Amplitude – the magnitude of the maximum
displacement (in either direction)
x(t )  xm cos(t   )
Phase
x(t )  xm cos(t   )
Phase constant
x(t )  xm cos(t   )
Angular frequency
x(t )  xm cos(t   )
 0
xm cos t  xm cos  (t  T )
2

cos   cos(  2 )

T
cos(t  2 )  cos  (t  T )
  2f
2  T
Period
x(t )  xm cos(t   )
T
2

Velocity of simple harmonic motion
x(t )  xm cos(t   )
dx (t )
v (t ) 
dt
d [ xm cos(t   )]

dt
v(t )  xm sin( t   )
Acceleration of simple harmonic motion
x(t )  xm cos(t   )
2
dv(t ) d x(t )
a(t ) 

2
dt
dt
2
  xm cos(t   )
a(t )   x(t )
2
The force law for simple harmonic
motion
• From the Newton’s Second Law:
2
F  ma  m x
• For simple harmonic motion, the force is
proportional to the displacement
• Hooke’s law:
F  kx
k  m
2
k

m
m
T  2
k
Chapter 15
Problem 16
Energy in simple harmonic motion
• Potential energy of a spring:
U (t )  kx / 2  (kx / 2) cos (t   )
2
m
2
2
• Kinetic energy of a mass:
K (t )  mv / 2  (m x / 2) sin (t   )
2
2
 (kx / 2) sin (t   )
2
m
2
2
m
2
m  k
2
Energy in simple harmonic motion
U (t )  K (t ) 
 (kx / 2) cos (t   )  (kx / 2) sin (t   )
2
m
2

2
m
2
 (kx / 2) cos (t   )  sin (t   )
2
m
 (kx / 2)
2
m
2
2

E  U  K  (kx / 2)
2
m
Chapter 15
Problem 37
Pendulums
• Simple pendulum:
• Restoring torque:
   L( Fg sin  )
• From the Newton’s Second Law:
I     L( Fg sin  )
• For small angles
sin   
mgL
 

I
Pendulums
• Simple pendulum:
at

L
s

L
mgL
 

I
mgL
a
s
I
• On the other hand
a(t )   x(t )
2
mgL

I
Pendulums
• Simple pendulum:
mgL

I
mgL


2
mL
2
I  mL
2
g
L
L
T
 2

g
Pendulums
• Physical pendulum:
mgh

I
2
I
T
 2

mgh
Chapter 15
Problem 51
Simple harmonic motion and uniform
circular motion
• Simple harmonic motion is the projection of uniform
circular motion on the diameter of the circle in which
the circular motion occurs
x(t )  xm cos(t   )
Simple harmonic motion and uniform
circular motion
• Simple harmonic motion is the projection of uniform
circular motion on the diameter of the circle in which
the circular motion occurs
x(t )  xm cos(t   )
dx(t )
vx (t ) 
dt
vx (t )  xm sin( t   )
Simple harmonic motion and uniform
circular motion
• Simple harmonic motion is the projection of uniform
circular motion on the diameter of the circle in which
the circular motion occurs
x(t )  xm cos(t   )
2
d x(t )
ax (t ) 
2
dt
2
a x (t )   xm cos(t   )
Damped simple harmonic motion
Fb  bv
Damping
force
Damping
constant
Forced oscillations and resonance
• Swinging without outside help – free oscillations
• Swinging with outside help – forced oscillations
• If ωd is a frequency of a driving force, then forced
oscillations can be described by:
x(t )  xm (d / , b) cos(d t   )
• Resonance:
d  
Answers to the even-numbered problems
Chapter 15:
Problem 2
(a) 10 N;
(b) 1.2 × 102 N/m
Answers to the even-numbered problems
Chapter 15:
Problem 28
(a) 200 N/m;
(b) 1.39 kg;
(c) 1.91 Hz
Answers to the even-numbered problems
Chapter 15:
Problem 38
12 s
Answers to the even-numbered problems
Chapter 15:
Problem 42
(a) 0.499 m;
(b) 0.940 mJ
Answers to the even-numbered problems
Chapter 15:
Problem 58
6.0%