Transcript 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 )  A cos(t   )
Amplitude
• Amplitude – the magnitude of the maximum
displacement (in either direction)
x(t )  A cos(t   )
Phase
x(t )  A cos(t   )
Phase constant
x(t )  A cos(t   )
Angular frequency
x(t )  A cos(t   )
 0
A cos t  A cos  (t  T )
cos   cos(  2 )
cos(t  2 )  cos  (t  T )
2  T
2

T
  2f
Period
x(t )  A cos(t   )
T
2

Velocity of simple harmonic motion
x(t )  A cos(t   )
dx (t )
v (t ) 
dt
d [ A cos(t   )]

dt
v(t )  A sin( t   )
Acceleration of simple harmonic motion
x(t )  A cos(t   )
2
dv(t ) d x(t )
a(t ) 

2
dt
dt
2
  A cos(t   )
a(t )   x(t )
2
Chapter 15
Problem 5
A particle moving along the x axis in simple harmonic motion starts from its
equilibrium position, the origin, at t = 0 and moves to the right. The amplitude
of its motion is 2.00 cm, and the frequency is 1.50 Hz. (a) Show that the position
of the particle is given by x = (2.00 cm) sin (3.00 π t). Determine (b) the
maximum speed and the earliest time (t > 0) at which the particle has this
speed, (c) the maximum acceleration and the earliest time (t > 0) at which the
particle has this acceleration, and (d) the total distance traveled between t = 0
and t = 1.00 s.
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
Energy in simple harmonic motion
• Potential energy of a spring:
U (t )  kx / 2  (kA / 2) cos (t   )
2
2
2
• Kinetic energy of a mass:
K (t )  mv / 2  (m A / 2) sin (t   )
2
2
 (kA / 2) sin (t   )
2
2
2
2
m  k
2
Energy in simple harmonic motion
U (t )  K (t ) 
 (kA / 2) cos (t   )  (kA / 2) sin (t   )
2
2

2
2
 (kA / 2) cos (t   )  sin (t   )
2
 (kA / 2)
2
2
2

E  U  K  (kA / 2)
2
Chapter 15
Problem 17
A 50.0-g object connected to a spring with a force constant of 35.0 N/m
oscillates on a horizontal, frictionless surface with an amplitude of 4.00 cm.
Find (a) the total energy of the system and (b) the speed of the object when the
position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy
when the position is 3.00 cm.
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 27
A particle of mass m slides without friction inside a hemispherical bowl of
radius R. Show that if the particle starts from rest with a small displacement
from equilibrium, it moves in simple harmonic motion with an angular
frequency equal to that of a simple pendulum of length R. That is,   g / R
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
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 )  A cos(t   )
dx(t )
v x (t ) 
dt
vx (t )  A 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 )  A cos(t   )
dx(t )
v x (t ) 
dt
vx (t )  A 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 )  A cos(t   )
2
d x(t )
a x (t ) 
2
dt
2
a x (t )   A 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 )  A(d / , b) cos(d t   )
• Resonance:
d  
Questions?
Answers to the even-numbered problems
Chapter 15
Problem 2
(a) 4.33 cm
(b) −5.00 cm/s
(c) −17.3 cm/s2
(d) 3.14 s; 5.00 cm
Answers to the even-numbered problems
Chapter 15
Problem 16
(a)0.153 J
(b) 0.784 m/s
(c) 17.5 m/s2
Answers to the even-numbered problems
Chapter 15
Problem 26
1.42 s; 0.499 m