Transcript PHYS 218 - Texas A&M University

PHYS 218
sec. 517-520
Review
Chap. 13
Periodic Motion
What you have to know
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Periodic motions (oscillations)
Physical quantities describing periodic motions
Equation of motion for simple harmonic motion
Various examples of harmonic motion
Pendulums
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We skip the parts which have an asterisk (*) in the textbook.
Physical quantities
Restoring force
Equilibrium point
Amplitude (A): maximum displacement from equilibrium ® the maximum value of x
Cycle: one complete round trip
Period (T ): the time for one cycle, unit s (sec)
Frequency ( f ): the number of cycles in unit time, unit Hz (= s- 1 )
Angular frequency (w): w = 2p f
Basic relations
f =
1
1
2p
, T = , w = 2p f =
T
f
T
Simple harmonic motion (SHM)
This is the simplest case when the restoring force is directly proportional to the
displacement from the equilibrium
In this case, the restoring force is in the form of Fx = - kx.
d 2x
Becasue F = ma = m 2 ,
dt
d 2x
k
=
x : This is the basic equation for SHM
2
dt
m
w does not depend on A
Its solution is sinusoidal so sine or cosine function;
x = A cos q = A cos (wt )
Therefore, the SHM can be described by
d 2x
2
=
w
x
2
dt
Then it follows that
f=
w
1
=
2p 2p
k
1
m
, T = = 2p
m
f
k
k
w=
m
for SHM
Ex 13.2
SHM
A spring: 6 N force causes a displacement of 0.03 m
Attach a 0.5 kg object to the end of the spring, pull it a distance of 0.02 m nad then release it
Spring constant
The first condition determines the spring constant
F = kx Þ k =
F
x
=
6.0 N
= 200 N/m = 200 kg/s 2
0.03 m
w, f, T of this SHM
Since m = 0.5 kg,
w=
f =
k
=
m
200 kg/s 2
= 20 rad/s
0.5 kg
w
20 rad/s
=
= 3.2 /s = 3.2 Hz,
2p
2p
T=
1
1
=
= 0.31 s
f
3.2 Hz
Displacement in SHM
d 2x
2
x
w
=
Equation of motion:
2
dt
The general solution of this equation is
x = A cos(wt + f )
A : amplitude, f : phase angle
The two constants A and f can be determined
by initial conditions such as the position and velocity at t = 0
At t = 0, x = x0 , where x0 = A cos f
velocity in SHM
dx d
= (A cos (wt + f )) = - Aw sin (wt + f )
dt dt
acceleration in SHM
vx =
d 2 x dv
ax = 2 =
= - Aw2 cos (wt + f ) = - w2 x
dt
dt
Therefore, the velocity oscillates between w A and - w A
and the acceleration oscillates between w2 A and - w2 A.
Displacement in SHM (2)
At t = 0, x = x0 and v = v0 , where x0 = A cos f and v0 = - w A sin f
2
\ x = A cos f and v = (w A) sin f Þ
2
0
2
2
2
0
2
x02
v02
1 = cos f + sin f = 2 + 2 2
A
w A
Therefore,
v02
A = x + 2 Þ
w
Also
2
2
0
A=
v02
x + 2
w
2
0
æ v0 ö
v0 - w A sin f
÷
÷
=
= - w tan f Þ f = arctan ççç÷
÷
x0
A cos f
è x0 ø
2
2
Energy in SHM
In SHM, x = A cos(wt + f ), v = - Aw sin (wt + f )
Total energy
1 2 1 2 1
1
mv + kx = mA2 w2 sin 2 (wt + f )+ kA2 cos 2 (wt + f )
2
2
2
2
1
k
= kA2 {sin 2 (wt + f )+ cos 2 (wt + f )}
¬ using w2 =
2
m
1
It should be so because of energy
constant
= kA2
2
conservation
E=
Since x contains a sine function and v has a cosine function,
when x = ± xmax , v = 0 and
Since the energy relation gives
when x = 0, v = ± vmax
1 2 1 2 1 2
kA = mv + kx ,
2
2
2
we get v = ±
k
A2 - x 2
m
Ex 13.4
Velocity, acceleration, and energy in SHM
k = 200 N/m, m = 0.5 kg and the oscillating mass is released from rest at x = 0.02 m.
Maximum and minimum velocity
As we derived in the previous page,
v= ±
k
A2 - x 2 .
m
Þ vmax =
vmin = -
k
A=
m
Note that the minimum value is NOT 0!
200 N/m
´ (0.02 m) = 0.40 m/s
0.5 kg
k
A = - 0.40 m/s
m
Maximum acceleration
k
x,
m
kA (200 N/m)(0.02 m)
=
=
= 8.0 m/s 2
m
0.50 kg
Since a = - w2 x = amax =
k
x
m max
Ex 13.4
Cont’d
The body has moved halfway to the center from its original position
The position of the body is x = A / 2 = 0.01 m.
Then,
k
v= A2 - x 2 = - 0.35 m/s
m
k
a= x = - 4.0 m/s 2
m
The body is moving from x=A/2 to x=0.
Thus we choose the negative sign for
the velocity.
Ex 13.5
Energy and momentum in SHM
This is a collision problem.
Þ always consider momentum conservation!
m
v1
M
O
To use momentum conservation,
we should first know the velocity of the body.
The velocity can be obtained by energy consideration.
Velocity at x=0
Before the collision, the total energy is E1 =
At x = 0, U = 0, and E1 =
1 2
kA1
2
1 2 1
kA1 = Mv12 Þ
2
2
Collision
By momentum conservation in x-direction,
M
M v1 + 0 = (M + m)v2 Þ v2 =
v1
M+m
v1 =
k
A1
M
The putty m is moving in y-direction.
Cont’d
Ex 13.5
After the Collision
Now the body (M + m) is moving with v2 at x = 0.
Then the total energy becomes
2
æ M
ö æ M ö÷
1
1
E2 = (M + m)v22 = (M + m)çç
v1 ÷
= çç
÷
÷E
÷
ç
è M + m ø èç M + m ø÷ 1
2
2
The total energy can also be written in terms of the amplitude as E =
therefore,
ö1 2
1 2 æ
M ÷
kA2 = çç
kA1 Þ
÷
÷
ç
è
ø
2
M+m 2
The period is
T = 2p
A2 = A1
mass
spring constant
Then the period after the collision is
T = 2p
M+m
k
M
M+m
1 2
kA ,
2
Cont’d
Ex 13.5
If the Collision happens when M is at x=A
The velocity of the block is then zero.
There is no motion and the collision just causes the change of
the mass from M to m.
Since the amplitude does not change, the
total energy of the system does not change.
The period depends on the mass, so
M+m
k
after the collision.
T = 2p
Vertical SHM
oscillation
equilibrium
l
l
l
Dl
Dl - x
F = kD l
F = mg
In equilibrium,
k D l = mg
F = kD l
x
F = mg
Fnet = k (D l - x)- mg = - kx
Therefore,
d 2x
k
=x Þ
dt
m
SHM
Angular SHM
q : angular displacement
Equation of motion (from Newton's 2nd law)
t = Ia
If the restoring torque is proportional to the angular displacement
t = - kq
Therefore, we get
d 2q
k
a = 2 = - q Þ SHM
dt
I
The angular frequency is w =
k
I
Simple pendulum
Equation of motion along the x direction
Fq = - mg sin q
q
(cf x = Lq)
Therefore,
L
d 2x
g
=
g
sin
q
»
g
q
=
x Þ SHM
2
dt
L
The angular frequency is
T
x
mg cos q
mg sin q
w=
g
L
mg
The Taylor expansion
When q is small
sin q ; q
Physical pendulum
Any real pendulum
pivot
q
You can treat this object as a point-like object which has the mass of the object
at its center-of-mass position.
Then it becomes similar to the case of simple pendulum.
d
t = - (mg )(d sin q) = I a
When q is small, sin q » q,
d 2q
mgd
Þ
=
q
2
dt
I
angular frequency
mg
w=
mgd
I
Þ
SHM
cf . For simple pendulum, I = md 2
then we have w =
mgd
=
2
md
g
d
Damped oscillations
Forced oscillations & resonance
These topics will not be summarized here.
But you should read the textbook for these topics.