Transcript Simple Harmonic Motion

Chapter 15
Oscillations
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Simple Harmonic Motion
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Oscillations
Let’s assume we have a horizontal spring and mass system
composed of a spring with the elastic (spring) constant k, and a
mass m.
• If we stretch the spring with
the mass on the end and let it
go, the mass will
continuously move back and
forth (if there is no friction).
k
k
• The back and forth or up and
down motion that repeats
itself on the same path at
equal intervals of time is
called oscillation (vibration).
k
m
m
m
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Characteristics of Oscillations
Horizontal
Spring
x -stretch- displacement
from the equilibrium
(unstretched) position.
•The motion will repeat itself
after an interval of time T
called period.
Maximum displacementamplitude A
•T is measured in seconds s.
The number of oscillations per unit of time is
called frequency f. It is measured in Hz or s-1.
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• If n oscillations are happening in the time t, then:
T = t/n, and f = n/t.
• Therefore T= 1/f,
f = 1/T, or
fT =1
Example T, f
• A mass oscillates 20 times in 5s. What is the period and
the frequency of the oscillation?
T = 5s/20 = 0.25 s
f = 20/5s = 4 Hz
Check Tf = 1,
4 Hz x .25 s = 1
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SHM Dynamics
At any given instant the
force F = -kx called
restoring force acts on
the mass.
It is proportional to the
displacement, opposes
motion and points towards
the equilibrium position.
F = -kx
k
a
m
x
An oscillation caused by a restoring force is
called Simple Harmonic Motion –SHM.
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SHM Dynamics
• At any given instant we know
that F = ma must be true.
F = -kx
• But in this case F = -kx
d x
dt 2
d 2x
m 2
dt
and ma = m
• So: -kx = ma =
2
d 2x
k


x
2
m
dt
k
a
m
x
a differential equation for x(t)!
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Solution for the
differential equation:
Where:
A = amplitude: max displacement from equilibrium position
= angular frequency
t + δ = phase
δ = phase constant (phase @ t =0)
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Comparison b/w two oscillations:
• Lets assume we have two oscillations:
x1  A cos t
and
x2  A cos(t   )
If δ = 2nπ, n= …-2,-1,0,1,2,… than:
x1=x2, and the systems are in phase.
If δ = (2n+1)π, n= …-2,-1,0,1,2,… than:
x1=-x2, and the systems are (180o) out of phase.
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Citicorp
building
in New
York.
Spring and 400 ton mass system oscillates out
of phase with building reducing the swaying
during high winds (same natural frequency).
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The derivative of x gives the velocity v:
Differentiating the velocity with respect to the time we
get the acceleration:
dv d 2 x
a
 2   2 A cos(t   )
dt
dt
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Since: x  A cos(t   ) ,
we get a relationship b/w the position and acceleration:
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Comparing a=-(k/m)x with a = -ω2x we get:
k

m
A and δ can be determined from the initial conditions xo and vo.
Setting t=0 in:
x  A cos(t   )
v  A sin( t   )
x0
 cos 
A
gives: v  A sin 
o
2
  vo 2
 2 2
  A
 
Also:

 1


2
 x0 
 v 
    o  1
 A
 A 
Square and add:
vo
 sin 
 A
 x0 2
 2
 A

x0  A cos 
A
x 
2
o
vo2
2
  arccos
xo
A
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The object is at the same location after one period T: x(t) = x(t+T)
x  A cos(t   ) A cos[ (t  T )   ]  A cos[t  T )   ]
Since cos repeats in value after 2π we must have:
T  2 or  
2
 2f
T
Solving for f:
or:
Or the period:
m
T  2
k
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Astronaut measures his mass
by sitting in a seat attached to
a spring and oscillating back
and forth. The total mass of
astronaut + seat is related to
the frequency of oscillation
by the equation:
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If δ = 0, the equations
for x, v, and a are:
x  A cos(t )
v  A sin( t )
a   2 A cos(t )
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Simple Harmonic Motion and Circular Motion
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Bubbles from a propeller rotating in water produce a
sinusoidal pattern.
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An object oscillates with ω = 8 rad/s. At t = 0, the object
is at xo = 4 cm, with the initial velocity of – 25 cm/s.
a) Find the amplitude and the phase constant for this motion.
Use
x0  A cos 
x  A cos(t   )
with t =0:
vo  A sin 
v  A sin( t   )
Rearranging:
vo
 sin 
 A
x0
 cos 
A
  tan
1
Divide
vo
 tan 
 xo
vo
 25cm / s
1
 tan (
)  0.663rad
 xo
8rad / s  4cm)
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x0
4cm
A

 5.08cm
cos 
cos 0.663rad
b) Write the law of motion:
x  (5.08cm) cos[(8s 1 )t  0.663]
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Energy in Simple
Harmonic Motion
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2
kx
U
2
mv 2
K
2
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Graphs of x, U, K vs. time
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Graph of U vs. x. The blue line is the total energy.
The kinetic energy is the vertical distance K=Etotal-U
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Some Oscillating Systems
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Object on Vertical Spring:
Unstretched:
mg
0   kyo  mg  yo 
k
Object oscillating:
Fy  ky  mg
Change variable: y=yo+y’
Fy  k ( yo  y)  mg
Fy  kyo  ky  mg
Fy  ky
d 2 y
m 2  ky
Standard SHM equation with solution:
dt
y  A cos(t   )
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Also potential energy:
2

ky
U
Uo
2
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Example 1
A small mass m1 rests on but is not
attached to a large mass M2 that
slides on its base without friction.
The maximum frictional force between
m1 and M2 is f. A spring of spring
constant k is attached to the large
mass M2 and to the wall as shown.
f
a. Determine the maximum horizontal acceleration that M2 may have without causing
m1 to slip.
amax = f/m1
b. Determine the maximum amplitude A for simple harmonic motion of the
two masses if they are to move together, i.e., m1 must not slip on M2.
In
d 2x
k


x
2
m
dt
m = m1 + M2
amax = ω2A
f
k
m1  M 2 f  M 2  f


A A
  1 
m1 m1  M 2
m1
k 
m1  k
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O
x
-kx
Example 1 Continuation
c. The two-mass combination is
pulled to the right the maximum
amplitude A found in part (b)
and released. Describe the
frictional force on the small
mass m1 during the first half
cycle of oscillation.
f
c) f = m1a, follows acceleration, decrease for T/4, than increase for T/4
d. The two-mass combination is now pulled to the right a distance of A'
greater than A and released.
i. Determine the acceleration of m1 at the instant the masses are released.
amax = ω2A’
a1  
k
A'
m1  M 2
ii. Determine the acceleration of M2 at the instant the masses are released.
f-kA’ = M2a2
k '
f
a2  
A
M2
M2
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Example 2
A block rests on a spring and oscillates
vertically with 4Hz, and amplitude 7 cm.
A tiny bead is placed on the top of the
oscillating block as it reaches its lowest
point. Assume that the bead’s mass is
negligible. At what distance from the
block’s equilibrium position will the bead
lose contact with the block?
N
mg
ay
-N + mg = may
Lose contact at N=0
ay = g
a y   2 y
  2f
g   2 y
g  (2f ) 2 y
g
9.81m / s 2
y

 0.0155m  1.55cm
(2f ) 2 [2 (4Hz )]2
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Simple Pendulum:
d 2s
 mg sin   m 2
dt
s  L
sin   
2
d 
 g  L 2
dt
d 2
g
 
2
dt
L
g
2
   
L
d 
2




2
dt
g 2

L T
2
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Pendulum in accelerated
frame of reference
g '  g 2  ao2
L
T  2
g'
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Physical Pendulum:
  I
d 2
 MgD sin   I 2
dt
MgD
d 2

sin   2
I
dt
MgD
d 2
sin   

 2
I
dt
MgD
 2
I
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A uniform rod of mass M and length L is
free to rotate about a horizontal axis
perpendicular to the rod and through one
end of the rod. a) Find the period of
oscillation for small angular displacements.
T  2
I
MgD
ML2
I
3
D = L/2
ML2
3  2 2 L
T  2
L
3g
Mg
2
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A uniform rod of mass M and length L is
free to rotate about a horizontal axis
perpendicular to the rod. b) Find the period
of oscillation if the rotation axis is a
distance x from the CM.
D=x
T  2
T  2
I  I CM
2
ML
 Mx 2 
 Mx 2
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I
MgD
ML2
 Mx 2
12
 2
Mgx
1
12
L2  x 2
gx
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Damped Oscillations
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Amplitude decreases in time
due to resistive forces
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Driven Oscillations and
Resonance
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