Transcript lec SHM teacher

SIMPLE HARMOIC MOTION
CCHS Physics
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Facts of SHM
• SHM occurs when an object is vibrating at a
single frequency and period
• PERIODOIC: when a vibration or oscillation
repeats itself, back and forth, over the same path
• We’ll deal mainly with simple
harmonic oscillations where
the position of the object
can be described by a
sinusoidal (sin, cos)
function
Examples of SHM
• Block attached to spring
• Motion of a pendulum
• Vibrations of a stringed musical
instrument
• Clocks
• Oscillations of houses, bridges, …
Meaning of
Simple Harmonic Motion
• SIMPLE  single frequency
• HARMONIC  sinusoidal
• A system in which the restoring force is
proportional to the negative
displacement (Hooke’s Law) will move
in simple harmonic motion.
Spring and SHM
Example of Complex
Harmonic Motion
Approximately simpleharmonic in this region
Hooke’s Law
• The magnitude of the restoring
force is proportional to the
displacement
F  kx
• Where:
– F = force (N)
– k = spring constant (N/m)
– x = displacement
• Acceleration
k
a x
m
• Force and Acceleration
– Proportional to x
– Directed toward the equilibrium
position
Hooke’s Law cont.
• Accurate as long as there is not too
much displacement
• There is a negative sign because the
force always acts opposite to the
displacement
• Note: F varies with position
Five Fun Definitions
• DISPLACEMENT: the distance x from the
equilibrium point
• AMPLITUDE: the maximum displacement
• CYCLE: one complete to-and-from motion
• PERIOD (T): time to complete one cycle
• FREQUENCY (f): number of cycles per
second
– Frequency is measured in s-1 = Hertz (Hz)
Period and Frequency
• Period and Frequency are inversely
related:
1
f 
T
1
T
f
Force vs. Distance Graphs
• Slope = spring constant
• Area = work (or energy)
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1
Area  bh
2
1
 x kx 
2
1
Work  kx 2  Energy
2
PEspring
x
1 2
 U s  kx
2
1 2
0 kxdx  2 kx
Energy of a Spring
• Remember that potential energy is stored
energy due to position
• The calculation of potential energy is
equivalent to calculating work
PEspring
1 2
 U s  kx
2
• Note: usually going to neglect mass of spring
Example 1
Given the following graph:
• What is the spring constant?
20
slope 
k  4 N/m
5
Force (N)
20
0
0
5
disp (m)
• If a 2 kg mass attached to this spring is displaced 3 m,
what is the acceleration of the mass?
kx 4 3
a

a  6 m/s2
m
2
• If the spring is stretched to 5 m, how much energy is
stored in the spring?
1 2 1
2
U s  kx  4 5 
U  50 J
2
2
Conservation of Energy
• Mechanical Energy - the sum of all kinetic
and potential energies
– Now we add in elastic potential energy
Etotal  KE  U g  U s
Etotal 
1 2
1
mv  mgh  kx 2
2
2
• We still have that energy must be
conserved, thus:
Einitial  E final
1 2
1 2
1 2
1 2
mv1  mgh1  kx1  WNC  mv2  mgh2  kx2
2
2
2
2
Example 2
• A spring stretches by 5 cm when a 100 g
mass is hung from it. A 250 g mass
attached to this spring and pulled back
0.70 m and released from rest. What is
the speed of the block as it passes
through the equilibrium point?
• First, find the spring constant:
F  kx
.19.8   k .05 
• Now, use conservation of energy
 k  19.6 N/m
1 2
1
1
1
mv1  mgh1  kx12  WNC  mv22  mgh2  kx22
2
2
2
2
1 2 1 2
kx1  mv2
2
2
1
1
2
19.6 .7   .250 v22
2
2
v  6.19 m/s
Example 3
•
A 0.2-kg ball is attached to a vertical massless spring as
shown. The spring constant of the spring is 28 N/m. The
ball, supported initially so that the spring is neither
stretched nor compressed, is released from rest. In the
absence of air resistance, how far does the ball fall
before being brought to a momentary stop by the spring?
• Conservation of Energy
1 2
1
1
1
mv1  mgh1  kx12  WNC  mv22  mgh2  kx22
2
2
2
2
1 2
mgh1  kx2
2
– Note: x2 = h1 call this distance d
1
.2 9.8 d  28 d 2
2
d  0.14 m
Velocity and Position
• Initially the mass is at its
maximum extension A
– Entirely elastic potential energy
• The initial elastic potential
energy converted to a
combination of kinetic and
eleastic potential energy
1 2 1 2 1 2
kA  mv2  kx2
2
2
2
• Solving for v

k 2
v
A  x2
m

Simple Harmonic Motion vs.
Uniform Circular Motion
• A ball is glued to the top of a turntable
• Focus on the shadow of the ball
• If the turntable rotates with constant angular velocity the
shadow of the ball moves in simple harmonic motion
SHM vs. UCM cont.
• If we can prove that the velocity of the shadow varies with position
like the function on the previous slides, we know the shadow moves
in SHM.
2
2
vC A x
•
Looking at the top triangle (made with
v
velocity vectors)
sin  
•
Now looking at the larger triangle
sin  
•
A2  x 2
A
Equating these two equations
v

v0
v0
v
A
•
v0
A2  x 2
A
A2  x 2  C A2  x 2
Thus we have proved that the velocity of the shadow in the x direction is
related to displacement in exactly the same manner as the velocity of an
object undergoing SHM.
Frequency and Period
• The velocity of the ball in the previous slide is:
or T  2 A
v
• Where 0
– A is the amplitude
– T is the period
T 2 A

4
4v0
2 A
v0 
T
• Focus on 1/4 of the trip,
• Imagine the shadow is a block on a spring, in the 1/4
of a cycle the block moves from a point of all elastic
potential energy to a point of all kinetic energy
A
1 2 1 2
kA  mv0  
v0
2
2
m
k
• Substituting for A/v0
m
Ts  2
k
Period of a Pendulum
• Restoring force = mgsin
• Not SHM because F  sin, not 
• However, for small :
sin ≈ (measured in radians)
• Thus, F = -mgsin ≈ mg
• Also, s = L
 mg 
Ft   
s

 L 
• Essentially SHM (Hooke’s Law)
• Plugging into period of a spring
formula with k = (mg)/L
m
m
T  2
 2
k
mg L
L
T p  2
g
Displacement Equation
• Looking at reference circle again
• Ball starts on the x axis at x = +A and
moves through the angle  in a time t
• Ball rotates at constant angular
speed  (because UCM)
  = t
• Displacement x of the shadow is just
the projection of the radius A onto the
x-axis
x  Acos  Acos  t 
Properties of SHM
Position of particle at time t:
x(t)  Acos( t   )
A…amplitude
…Angular frequency
…phase constant, phase angle
T…period
t F…phase
x(t )  A cos(t )
Period and Frequency (Revisited)
• PERIOD: time required to complete one cycle
• The value of T depends on the angular speed
(frequency) 
– Greater , less T
 angular displacement


t
time
2

T
• For 1 cycle,  = 2, and t = T:
• Using the fact that frequency and period are
inverses:   2 f
2

 2 f
T
Velocity and Acceleration
x  A cos( t   )
v   Asin( t   )
a   A cos( t   )
2
dx
dv
note: v t  
and a t  
dt
dt
More Properties of SHM
xmax  A
vmax   A
amax   2 A
Phase of velocity differs by – /2 or 90° from phase of displacement.
Phase of acceleration differs by –  or 180° from phase of displacement.
Even More Properties of SHM
• Acceleration of particle is
proportional to the displacement, but
is in the opposite direction
(a = - 2·x).
• Displacement, velocity and
acceleration vary sinusoidally.
• The frequency and period of the
motion are independent of the
amplitude.
Summary of SHM
t
x
v
a
KE
Us
0
A
0
-2A
0
½kA2
T/4
0
-A
0
½kA2
0
T/2
-A
0
-2A
0
½kA2
3T/4
0
-A
0
½kA2
0
T
A
0
-2A
0
½kA2
Example
a)
A block whose mass m is 680 g is fastened to a spring
whose spring constant k is 65 N/m. The block is pulled a
distance of x = 11 cm from its equilibrium position at x = 0
on a frictionless surface and released from rest at t = 0.
What force does the spring exert on the block just before
the block is released?
F  kx   65.11  7.2N
b)
What are the period, frequency and angular frequency?
m
.680
Ts  2
 2
 0.64 s
k
65
1
1
f  
 1.56 Hz
T .64
  2 f  2 1.56  9.78 rad/s
Example cont.
c)
What is the amplitude of the oscillation?
The amplitude is 11 cm (that is where it is released from)
d)
What is the maximum speed of the oscillating block?
vmax   A  9.78 .11  1.1 m/s
max speed occurs whenever x = 0
e)
What is the magnitude of the maximum acceleration of the
block?
amax   2 A  9.78 2 .11  11 m/s 2
max acceleration occurs at the end points
Example cont.
f)
What is the phase constant  of the motion?
At t = 0, the displacement is at its maximum value A, and
the velocity of the block is zero. If we put these initial
conditions into the displacement and velocity equations
we find:
x  A cos( t   )
.11  .11cos  0    
1  cos  
v   Asin( t   )
0  1.1sin  0    
0  sin  
The smallest angle that solves these two equations is
 = 0. (Any multiple of 2 will also work)