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Transcript
Oscillations
Unit 7
Lesson 1 : Simple Harmonic Motion
Fs = -kx
(Hooke’s Law)
Fs is a restoring force
because it always
points toward the
equilibrium position
(x = 0)
Applying Newton’s Second Law :
SF = max
-kx = max
k
ax = x
m
Simple Harmonic Motion
An object moves with simple harmonic motion
whenever its acceleration is proportional to its
position and is oppositely directed to the
displacement from equilibrium.
Example 1
A block on the end of a spring is pulled
to position x = A and released. In one full
cycle of its motion, through what total
distance does it travel ?
_____ A/2
_____ A
_____ 2A
_____ 4A
dv
dt
Since a =
d2x
dt2
= -
d2x
,
=
dt2
k
m
x
Notice that the
acceleration of the particle
in SHM is not constant. It
varies with position x.
If we call k/m = w2,
d2x
dt2
= -w2x
d2x
2x
=
-w
dt2
(second-order differential equation)
We need a function x(t) whose second
derivative is the same as the original function
with a negative sign and multiplied by w2.
One possible solution is :
x(t) = A cos(wt + f)
Proof
dx
d
= A
cos(wt + f) = -wA sin(wt + f)
dt
dt
d2x
d
2A cos(wt + f)
=
-wA
sin(wt
+
f)
=
-w
dt2
dt
d2x
dt2
= -w2x
x(t) = A cos(wt + f)
Amplitude (A) : maximum value of the
position of the particle in either the
positive or negative direction.
Angular Frequency (w) : number of
oscillations per second.
Since k/m = w2,
w=
k
m
Phase Constant (f) : initial phase angle.
This is determined by the position of
the particle at t = 0.
If particle is at maximum
position x = A at t = 0, the
phase constant f = 0.
f=0
Pen traces out cosine curve
x(t) = A cos(wt + f)
Period (T) : the time interval required for
the particle to go through one full
cycle of its motion.
T=
2p
w
Frequency (f) : the inverse of period. The
number of oscillations that the
particle undergoes per second.
f=
1
T
Since T = 2p/w,
f=
w
2p
The Hertz (Hz) is the SI unit for frequency.
Since f = w/2p,
2p
w = 2pf =
T
Since w =
depend only
on mass and
spring
constant
k/m and T = 2p/w ,
T = 2p
f=
1
2p
m
k
k
m
Velocity in Simple Harmonic Motion
dx
d
= A
cos(wt + f) = -wA sin(wt + f)
dt
dt
v = -wA sin(wt + f)
Acceleration in Simple Harmonic Motion
d2x
d
2A cos(wt + f)
=
-wA
sin(wt
+
f)
=
-w
dt2
dt
a = -w2A cos(wt + f)
Maximum Speed in Simple Harmonic Motion
Since sine and cosine oscillate between +/- 1,
vmax = +/- wA
vmax =
k
m
A
(magnitude only)
Maximum Acc. in Simple Harmonic Motion
amax = +/- w2A
amax =
k
m
A
(magnitude only)
position vs. time
phase difference
is p/2 rad or 90o
when x at max or min, v = 0
when x = 0, v is max
velocity vs. time
phase difference
is p rad or 180o
acceleration vs. time
when x at max, a is max in
opposite direction
Example 2
An object oscillates with simple harmonic
motion along the x-axis. Its position varies
with time according to the equation
x = (4.00 m) cos(pt + p/4)
where t is in seconds and the angles in the
parentheses are in radians.
a) Determine the amplitude, frequency,
and period of the motion.
b) Calculate the velocity and acceleration
of the object at any time t.
c) Using the results of part b, determine
the position, velocity, and acceleration
of the object at t = 1.00 s.
d) Determine the maximum speed and
maximum acceleration of the object.
e) Find the displacement of the object
between t = 0 and t = 1.00 s.
Example 3
A 200 g block connected to a light spring for
which the force constant is 5.00 N/m is free to
oscillate on a horizontal, frictionless surface. The
block is displaced 5.00 cm from equilibrium and
released from rest, as shown below.
a) Find the period of its motion.
b) Determine the maximum speed of the block.
c) What is the maximum acceleration of the
block ?
d) Express the position, speed, and acceleration
as functions of time.
e) The block is released from the same initial
position, xi = 5.00 cm, but with an initial
velocity of vi = -0.100 m/s. Which parts of
the solution change and what are the new
answers for those that do change ?
Example 4 : AP 1989 # 3
A 2 kg block is dropped from a height of 0.45 m
above an uncompressed spring, as shown above.
The spring has an elastic constant of 200 N/m and
negligible mass. The block strikes the end of the
spring and sticks to it.
a) Determine the speed of the block at the instant
it hits the end of the spring.
b) Determine the period of the simple harmonic
motion that ensues.
c) Determine the distance that the spring is
compressed at the instant the speed of the
block is maximum.
d) Determine the maximum compression of the
spring.
e) Determine the amplitude of the simple
harmonic motion.
Example 5 : AP 2003 # 2
An ideal spring is hung from the ceiling and a
pan of mass M is suspended from the end of the
spring, stretching it a distance D as shown
above. A piece of clay, also of mass M, is then
dropped from a height H onto the pan and sticks
to it. Express all algebraic answers in terms of
the given quantities and fundamental constants.
a) Determine the speed of the clay at the instant
it hits the pan.
b) Determine the speed of the pan just after the
clay strikes it.
c) Determine the period of the simple harmonic
motion that ensues.
d) Determine the distance the spring is stretched
(from its initial unstretched length) at the
moment the speed of the pan is a
maximum. Justify your answer.
The clay is now removed from the pan and the pan
is returned to equilibrium at the end of the spring.
A rubber ball, also of mass M, is dropped from the
same height H onto the pan, and after the collision
is caught in midair before hitting anything else.
e) Indicate below whether the period of the
resulting simple harmonic motion of the pan
is greater than, less than, or the same as it
was in part c.
____ Greater than
____ Less than
Justify your answer.
____ The same as
Lesson 2 : Energy in Simple Harmonic Motion
KE of Block
KE = ½ mv2
Since v = -wA sin(wt + f),
KE = ½ mw2A2 sin2(wt + f)
Elastic PE Energy Stored
in Spring
U = ½ kx2
Since x = A cos(wt + f),
U = ½ kA2 cos2(wt + f)
Total Mechanical Energy of
Simple Harmonic Oscillator
E = KE + U
E = ½ mw2A2 sin2(wt + f) + ½ kA2 cos2(wt + f)
Since w2 = k/m,
E = ½ kA2 [sin2(wt + f) + cos2(wt + f)]
Since sin2q + cos2q = 1,
E = ½ kA2
E = ½ kA2
The total mechanical energy of a simple
harmonic oscillator is a constant of the
motion and is proportional to the square
of the amplitude.
U is small when KE is
large, and vice versa.
KE + U = constant
E = ½ kA2
Since E = KE + U,
½ kA2 = ½ mv2 + ½ kx2
Solving for v,
v = +/-
k (A2 – x2)
m
Since w =
v = +/- w
k
m
A2 – x2
,
velocity of
simple
harmonic
oscillator
Example 1
The amplitude of a system moving in
simple harmonic motion is doubled.
Determine the change in the
a) total energy
b) maximum speed
c) maximum acceleration
d) period
Example 2
A 0.500 kg cart connected to a light
spring for which the force constant is
20.0 N/m oscillates on a horizontal,
frictionless air track.
a) Calculate the total energy of the system
and the maximum speed of the cart if
the amplitude of the motion is 3.00 cm.
b) What is the velocity of the cart when the
position is 2.00 cm ?
c) Compare the kinetic and potential
energies of the system when the
position is 2.00 cm ?
Lesson 3 : Comparing Simple Harmonic
Motion with Uniform Circular Motion
As the turntable rotates
with constant angular
speed, the shadow of
the ball moves back
and forth in simple
harmonic motion.
Reference Circle
Simple harmonic motion along a
straight line can be represented by the
projection of uniform circular motion
along a diameter of a reference circle.
t=0
Since q = wt = f ,
x coordinate is x(t) = A cos(wt + f)
Point Q moves
with simple
harmonic motion
t>0
Q is the
projection
of P
Example 1
What is the amplitude and
phase constant (relative to
an x axis to the right) of the
simple harmonic motion of
the ball’s shadow ?
_____ 0.50 m and 0
_____ 1.00 m and 0
_____ 0.50 m and p
_____ 1.00 m and p
Example 2
While riding behind a car
traveling at 3.00 m/s, you
notice that one of the car’s
tires has a small
hemispherical bump on its
rim, as shown.
a) Explain why the bump,
from your viewpoint
behind the car,
executes simple
harmonic motion.
b) If the radii of the car’s tires are 0.300 m,
what is the bump’s period of oscillation ?
Lesson 4 : The Pendulum
A simple pendulum
exhibits periodic motion.
It consists of a particlelike bob of mass m
suspended by a light
string of length L that is
fixed at the upper end.
Forces acting on bob
Tension in string
Gravitational force mg
Ft = mg sinq always
acts opposite to
the displacement
of the bob
Ft is a restoring force
d2s
Ft = -mg sinq = m
dt2
Since s = Lq and L is constant,
d2q
g
=dt2
L
sinq
for small values of q
d2q
g
=dt2
L
sinq
same form
d2x
dt2
= -
k
m
x
For small amplitudes (q < about 10o), the
motion of a pendulum is close to simple
harmonic motion.
Instead of x = A cos(wt + f),
q = qmax cos(wt + f)
Instead of w =
angular
frequency
for a simple
pendulum
w=
T=
2p
w
k
m
g
L
= 2p
L
g
,
w=
g
L
T = 2p
L
g
The period and frequency of a simple
pendulum depend only on the length of the
string and the acceleration due to gravity.
(Period and frequency are independent of mass.)
Example 1
Christian Huygens (1629-1695), the greatest
clockmaker in history, suggested that an
international unit of length could be defined
as the length of a simple pendulum having a
period of exactly 1s.
a) How much shorter would our length unit
be had his suggestion been followed ?
b) What if Huygens had been born on
another planet ? What would the value
of g have to be on that planet such that
the meter based on Huygen’s
pendulum would have the same value
as our meter ?
Example 2
A simple pendulum has a mass of 0.250 kg
and a length of 1.00 m. It is displaced
through an angle of 15.0o and released.
What is the
a) maximum speed
b) maximum angular acceleration
c) maximum restoring force ?
Example 3
A simple pendulum is 5.00 m long.
a) What is the period of small oscillations for this
pendulum if it is located in an elevator
accelerating upward at 5.00 m/s2 ?
b) What is its period if the elevator is accelerating
downward at 5.00 m/s2 ?
c) What is the period of this pendulum if it is
placed in a truck that is accelerating horizontally
at 5.00 m/s2 ?
The Physical Pendulum
A hanging object that oscillates about a
fixed axis that does not pass through its
center of mass and the object cannot be
approximated as a point mass.
Gravitational force
produces a torque about
an axis through O.
t = mgd sinq
Since St = Ia,
- mgd sinq = I
d2q
dt2
- mgd sinq = I
d2q
dt2
negative sign because t
tends to decrease q
(restoring force)
If q is small so that sinq is almost q,
d2q
dt2
= -(
mgd
I
)q
= - w2q
simple harmonic
motion equation
w=
mgd
I
Since the period of a pendulum is
T=
T = 2p
Period of a Physical
Pendulum
2p
,
w
I
mgd
Example 4
A uniform rod of mass M and
length L is pivoted about one
end and oscillates in a vertical
plane. Find the period of
oscillation if the amplitude of
the motion is small.
Torsional Pendulum
A rigid object suspended by
a wire. When the object is
twisted, the twisted wire
exerts a restoring torque
that is proportional to the
angular position.
t = -kq
k is called the torsion
constant of the wire
Since St = I
d2q
dt2
d2q
dt2
= -
,
k
q
I
simple harmonic
motion equation
w=
Period of a Torsional
Pendulum
k
I
T = 2p
I
k
Example 5
A torsional pendulum is formed by taking a
meter stick of mass 2.00 kg, and attaching to
its center a wire. With its upper end clamped,
the vertical wire supports the stick as the
stick turns in a horizontal plane. If the
resulting period is 3.00 minutes, what is the
torsion constant for the wire ?