= 121
K mv = m 2 A2 sin 2 ( t + )
2
2
1212
2
U = kx = kA cos ( t + )
2
2
12
E = K + U = kA sin 2 ( t + ) + cos2 ( t + )
2
12
E = kA
2
[
]
The total mechanical energy of a s.h.o. is a
constant of the motion and is proportional to
the square of the amplitude.
At x = A , E = max U stored in the spring.
At x = 0, E = max K
Energy is continuously being transformed betwn U
stored in the spring and K of the block.
1 21212
E = K + U = mv + kx = kA
2
2
2
k2
2
2
2
(A x ) = (A x )
v=
m
v is max at x = 0 and 0 at x = A
A projection of the circular
motion of a rotating ball
matches the SHM of an
object on a spring
Uniform circular motion
projected onto one
dimension is simple
harmonic motion.
A particle in uniform circular motion with radius A and
angular velocity .
When the particle is at angle ,
x = Acos
If the particle starts from 0 = 0
at t = 0, its angle at a later time t
is = t , where is the
particles angular velocity.
The particles x-component can
be expressed as
x(t) = A cos(t + )
The x - component of a
particle in uniform circular
motion is SHM
Fig. 15-14, p. 430
A particle rotates counterclockwise in a circle of radius 3.00m
with a constant angular speed of 8.00 rad/s. At t = 0, the
particle has an x coordinate of 2.00 m and is moving to the
right. (a) Determine the x-coordinate of the particle as a
function of time. (b) Find the x component of the particles
velocity and acceleration at any time t.
x = A cos( t + ) = (3.00m) cos(8.00t + )
2.00m = (3.00m ) cos(0 + )
= cos1
2.00m
0
= 48.2 = 0.841 rad
3.00m
x = (3.00m) cos(8.00t 0.841)
dx
= ( 3.00m)(8.00rad / s) sin(8.00t 0.841)
dt
= ( 24.0m / s ) sin(8.00t 0.841)
dv
a x = x = ( 24.0m / s )(8.00rad / s ) cos(8.00t 0.841)
dt
= (192m / s ) cos(8.00t 0.841)
vx =
Pendulum
d 2s
Ft = mg sin = m 2
dt
d 2
g
s = L
= sin
2
dt
L
g
small
=
L
=
g
L
2
L
T=
= 2
g
The period and frequency depend only on the length of
string and acceleration due to gravity.
Fig. 15-16, p. 432
Damped Oscillations
Nonconservative force retards the motion.
Mechanical energy of the system in time
and motion is damped.
r
r
R = bv
Retarding force
b: damping coefficient
F
x
r
r
R bv
= = kx bv x = ma x
d 2x
dx
kx b = m 2
dt
dt
x = Ae ( b 2 m ) t cos( t + )
2
=
0 =
k b
b
= 0
m 2m
2m
k
m
: natural
2
frequency
Fig. 15-20, p. 436
Damped oscillator
When the retarding force is small, the oscillatory
character of the motion is preserved but the
amplitude decreases in time, with the result that the
motion ultimately ceases.
Amplitude decays exponentially with time.
Oscillations dampen more rapidly for larger
retarding force.
overdamped
critically damped
underdamped
Magnitude of the retarding force small,
underdamped
b/2m < 0,
As b, amplitude of the oscillations
more rapidly.
When b bc such that bc/2m = 0,
system does not oscillate,
critically damped
If b/2m > 0, system is overdamped
What is the period of the motion of a damped oscillator given
that m = 250g, k = 85N/m, and b = 70g/s. How long does it
take for the amplitude of the damped oscillations to drop to
half its initial value?
m
0.25kg
T = 2
= 2
= 0.34 s
k
85 N / m
1
1
bt / 2 m
xm e
= xm
ln = ln(e bt / 2 m ) = bt / 2m
2
2
2m ln 1 2 ( 2)(0.25kg )(ln 1 / 2)
t=
=
= 5.0 s
b
0.070kg / s
Forced Oscillations
energy added periodically to maintain oscillatory behaviour
F ( t ) = F0 sin t
: angular frequency of driving force
dx
d 2x
F = ma F0 sin t b dt kx = m dt 2
x = A cos( t + )
F0 / m
A=
steady-state
2
b
2
22
( 0 ) +
m
0 =
k
m
natural frequency of undamped oscillator
For small damping, A is large
when 0
Resonance
0 resonance frequency
r
r
F is in phase with v
rr
Power delivered to oscillator, F v
max when in phase
A with damping (b 0),
and resonance curve broadens as b .
b = 0, steady state amplitude infinity as 0
A block weighing 40.0 N is suspended from a spring that has
a force constant of 200 N/m. The system is undamped and is
subjected to a harmonic driving force of frequency 10.0 Hz,
resulting in a force-motion amplitude of 2.00 cm. Determine
the maximum value of the driving force.
A=
F0 m
(
2
)
22
0
(
= 2 f = 20.0 s 1
(
2
F0 = mA 2 0
(
)
2
0 =
)
)
200
k
= 40.0 = 49.0 s 2
m
9.80
()
40.0
2
F0 =
2.00 10 ( 3 950 49.0) = 318 N
9.80