Transcript Chapter 15– Oscillations

Chapter 15– Oscillations
I.
Simple harmonic motion (SHM)
- Velocity
- Acceleration
II. Force law for SHM
- Simple linear harmonic oscillator
(block attached to spring moving in 1D)
III. Energy in SHM
IV. Angular harmonic oscillator
- Simple angular harmonic oscillator
(torsion pendulum)
V. Pendulum
VI. SHM and uniform circular motion
VII. Damped simple harmonic motion
VIII. Forced oscillations and resonance
I. Simple harmonic motion
Examples: boats, guitar strings, diaphragms in telephones,
oscillations of air molecules that transmit the sensation of
sound, oscillations of atoms in a solid that convey the
sensation of temperature.
In the “real world” the oscillations are damped 
motion dies out gradually 
Mechanical energy transferred to thermal (friction)
I. Simple harmonic motion
- Frequency:
Number of oscillations completed each second.
1
f 
T
Units: 1 Hertz = 1 oscillation per second = 1s-1
T = oscillation period  time for a complete oscillation (cycle).
- Periodic / harmonic motion:
Any motion that repeats itself in a particular way.
Simple Harmonic Motion (SHM) Periodic motion  sinusoidal
(sine or cosine) function of time.
Displacement:
Amplitude (xm):
x(t )  xm cos(t   )
xm, ω, φ =const.
Magnitude of the maximum displacement.
Phase of the motion: (ωt + φ)
Phase constant (φ):
Angle that depends on the particle’s
displacement and velocity at t=0.
- Periodic / harmonic motion:
Any motion that repeats itself in a particular way.
x(t )  xm cos(t   )
Particle oscillating  maximum speed at x=0
minimum speed at x= ± xm
Angular frequency (ω):
After one period  x(T+t)=x(t)
2

 2f
T
Units:
rad/s
xm cos t  xm cos  (t  T )
The cosine function first repeat itself after 2π = 1 period
 (t  T )  t  2  T  2
Velocity of SHM:
dx(t ) d
v(t ) 
 xm cos(t   )  xm sin( t   )
dt
dt
Velocity amplitude (vm): vm= ω·xm
t
Acceleration of SHM:
d 2 x(t ) dv(t ) d
2


a(t ) 




x
sin(

t


)



xm cos(t   )
m
2
dt
dt
dt
Acceleration amplitude (am): am= ω2·xm
a(t )   2 xm cos(t   )   2 x(t )
In SHM, the acceleration is proportional to the displacement but opposite
in sign and the two quantities are related by the square of the angular
frequency.
Different amplitude
t
Different period
t
Different phase angle
II. Force law for SHM
F  ma  (m 2 ) x
Restoring force similar to:
F  kx
Spring constant: mω2
SHM is the motion executed by a particle of mass m subject to a
force that is proportional to the displacement of the particle but
opposite in sign.
Linear simple harmonic oscillator
Angular frequency:
k

m
Period:
T
2

 2
m
k
III. Energy in SHM
Linear oscillator  ΔEmec= cte  back and forth
energy transfer between kinetic and potential energy.
Potential energy:
U (t ) 
1 2 1 2
kx  kxm cos 2 (t   )
2
2
Kinetic energy:
K (t ) 
1 2 1
1
mv  m 2 xm2 sin 2 (t   )   2  k / m  K (t )  kxm2 sin 2 (t   )
2
2
2
Mechanical energy:
1 2
1
1
kxm cos 2 (t   )  kxm2 sin 2 (t   )  kxm2 [sin 2 (t   )  cos 2 (t   )] 
2
2
2
1
E  U  K  kxm2
2
E U  K 
Mechanical energy:
1 2
1
1
kxm cos 2 (t   )  kxm2 sin 2 (t   )  kxm2 [sin 2 (t   )  cos 2 (t   )] 
2
2
2
1
E  U  K  kxm2
2
E U  K 
IV. Angular Simple Harmonic oscillator
If we rotate the disk some angular displacement
θ from its rest position it will oscillate about that
position in angular simple harmonic motion.
  
Restoring torque:
Angular form of Hooke’s law
Torsion constant: κ (Greek letter kappa)
(depends on length, diameter and material
of suspension wire)
Period:
T  2
I

Angular simple harmonic oscillator
Torsion pendulum
Linear simple harmonic motion
Angular simple harmonic motion
Linear acceleration: a = -ω2 x
Angular acceleration: α = - ω2 θ
Angular acceleration proportional to angular displacement from the
equilibrium position but tends to rotate the system in the opposite
direction to the displacement.
Force: F = -k x
Torque:
  
Torque is proportional to the angular displacement but tends to rotate
the system in the opposite direction.
V. Pendulum
Simple pendulum
Particle of mass “m” (pendulum’s
bob) suspended from one end of
an unstretchable massless string
of length “L” that is fixed at the
other end.
Restoring torque about pendulum’s pivot point:
  r F   L( Fg sin  )   L(mg sin  )  I
Equilibrium position: θ = 0
If θ is small  θ = sin θ
“I” is the pendulum’s rotational inertia about the
pivot point

 Lmg

I
SHM  α ~ - (cte)·θ
In figure above, as the pendulum moves to right its acceleration to the left
increases until it stops and starts moving to the left  The movement of a simple
pendulum swinging through only small angles is approximately SHM.
Simple pendulum angular acceleration: α = - ω2 θ = - (Lmg/I)· θ
T  2
I
L
 2
mgL
g
mgL 2


I
T
Small oscillation amplitude
I =m L2 rotational inertia of pendulum = simple pendulum  particle (bob) at
radius L from pivot.
Complicated distribution of mass
Physical pendulum:
about the pivot
Difference with respect to simple pendulum 
the restoring force component Fg sinθ has a
moment arm of distance “h” about the pivot point
rather than the string length L.
T  2
I
mgh
“I” depends on shape of physical pendulum
about O
A physical pendulum will not swing if it pivots at its COM  h=0 
T infinity  such pendulum will never complete one swing.
Physical pendulum oscillating about O with period T can be represented
by simple pendulum of length L0 and same period.
Center of oscillation: point along the physical pendulum at distance L0
from O.
A physical pendulum can be used to measure “g”.
Example : pendulum = rod of length L suspended
from one end.
h=L/2 (COM)
I (COM)=ML2 /12
I = I (COM)+M h2 = ML2 /12 + M (L/2)2 = ML2/3
I
(2 ) 2 I
4 2 ML2
8 2 L
T  2
g
 2
g
2
Mgh
Mh T 3M ( L / 2)
T
3T 2
VI. Simple Harmonic Motion and Uniform Circular Motion
Simple harmonic motion is the projection of uniform circular motion on a
diameter of the circle in which the later motion occurs.
Particle P’  uniform circular motion  φ’= ωt + φ
Particle P  projection of
particle P’ onto x-axis
 x(t)= xm cos (ωt + φ)
v = ω·R v = ω·xm
Projection of velocity of reference
particle = velocity SHM
ar = ω2·R a = ω2·xm
v(t )    xm sin( t   )
a(t )   2  xm cos(t   )
Projection of radial acceleration of reference particle =
acceleration SHM
VII. Damped Simple Harmonic Motion
Damped motion : when the motion of an oscillator is reduced by an
external force.
Example: Block oscillates vertically on a spring. From block, a rod
extends to a vane submerged in liquid.
Liquid exerts a damping force opposed to the motion
Fd  bv
b = damping constant (kg/s)
Assumption:
Gravitational force on block is negligible compared to the damping force
and the force on the block from the spring.
d 2x
dx
 bv  kx  ma  m 2  b  kx  0
dt
dt
Solution:
x(t )  xmebt / 2m cos(' t   )
Amplitude : xm e
 bt / 2 m
Frequency of damped
oscillator:
Displacement of
damped oscillator
Amplitude decreases exponentially with
time
k
b2
' 

m 4m 2
If b=0
(no damping)
' 
k

m
Mechanical energy 
Un-damped
oscillator:
1 2
E (t )  kxm
2
Damped
oscillator:
1 2 bt / m
E (t )  kxm e
2
VIII. Forced oscillations and resonance
Examples:
1) Person swinging in a swing without anyone pushing  Free oscillation
2) Someone pushes the swing periodically  Forced or driven oscillations
Driven oscillations
Two frequencies:
(1) Natural angular frequency ω of the system
 when system oscillates freely after
a sudden disturbance.
(2) External frequency ωd of the system 
angular frequency of the external driving force
causing the driven oscillations.
If “rigid support” moves up and down  forced
simple harmonic oscillator
Xm depends on ωd and ω
x(t )  xm cos(d t   )
Velocity of oscillations greatest when: ωd = ω
 Resonance  Displacement amplitude
greatest
An oscillator consists of a block of mass
0.500 kg connected to a spring. When set
into oscillation with amplitude 35.0 cm, the
oscillator repeats its motion every 0.500 s.
Find the (a) period, (b) frequency, (c)
angular frequency, (d) spring constant, (e)
maximum speed, and (f) magnitude of the
maximum force on the block from the
spring.
Solution
• (a) The motion repeats every 0.500 s so the period must
be T = 0.500 s.
• (b) The frequency is the reciprocal of the period:
• f = 1/T = 1/(0.500 s) = 2.00 Hz.
• (c) The angular frequency ω is ω = 2πf = 2π(2.00 Hz) =
12.6 rad/s.
• (d) The angular frequency is related to the spring
constant k and the mass m by . We solve for k:
• k = mω2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m.
• (e) Let xm be the amplitude. The maximum speed is
• vm = ωxm = (12.6 rad/s)(0.350 m) = 4.40 m/s.
• (f) The maximum force is exerted when the displacement
is a maximum and its magnitude is given by Fm = kxm =
(79.0 N/m)(0.350 m) = 27.6 N.
In the figure below, two identical springs of
spring constant 7580 N/m are attached to
a block of mass 0.245 kg. The block is set
oscillating on the frictionless floor. What is
the frequency of oscillation?
Solution
• When displaced from equilibrium, the net force
exerted by the springs is –2kx acting in a
direction so as to return the block to its
equilibrium position (x = 0). Since the
acceleration a = d2x/dt2, Newton’s second law
d x
yields
m
= 2 kx.
2
dt 2
• Substituting x = xm cos(ωt + φ) and simplifying,
2k
we find
2
 =
f =

1
=
2 2
m
2k
1

m 2
2(7580 N/m)
 39.6 Hz.
0.245 kg
A block weighing 14.0 N, which can slide without friction on
an incline at angle θ = 40.0°, is connected to the top of
the incline by a massless spring of unstretched length
0.450 m and spring constant 120 N/m. (a) How far from
the top of the incline is the block's equilibrium point? (b)
If the block is pulled slightly down the incline and
released, what is the period of the resulting oscillations?
Solution
• (a) the equilibrium position when the block is gently
lowered until forces balance. If the amount the spring is
stretched is x, then we examine force-components along
the incline surface and find
kx  mg sin   x 
14.0sin 40.0
 0.0750 m
120
• at equilibrium, The distance from the top of the incline is
therefore (0.450 + 0.075) m = 0.525 m.
• (b) Just as with a vertical spring, the effect of gravity (or
one of its components) is simply to shift the equilibrium
position; it does not change the characteristics (such as
the period) of simple harmonic motion.
14.0 9.80
T = 2
= 0.686 s.
120