Hooke`s Law Simple Harmonic Motion

Transcription

12/9/09
 
Hooke’s Law
 
Spring
 
Simple Harmonic Motion
 
Energy
Physics 201, UW-Madison
1
relaxed position
FX = -kx > 0
FX = 0
x0 x>0
x=0
x=0
12/9/09
FX = - kx < 0
xx
2
 
 
We know that if we stretch
a spring with a mass on
the end and let it go, the
mass will oscillate back
and forth (if there is no
friction).
This oscillation is called
Simple Harmonic
Motion, and is actually
easy to understand...
k
k
k
12/9/09
m
m
m
3
k
d2x
m 2
dt
x
d2x
k
=− x
2
dt
m
12/9/09
m
a differential equation
4
d2x
k
=− x
2
dt
m
define
d2x
2
=
−
ω
x
2
dt
ω=
k
m
Where ω is the angular
frequency of motion
Try the solution x = A cos(ωt)
dx
= −ω Asin (ω t )
dt
d2x
2
2
=
−
ω
A
cos
ω
t
=
−
ω
x
(
)
2
dt
This works, so it must be a solution!
12/9/09
5
 
We just showed that
d2x
2
=
−
ω
x
2
dt
(from F = ma)
has the solution x = A cos(ωt) but … x = A sin(ωt) is also a solution.
 
The most general solution is a linear combination (sum) of these two
solutions!
x = Bsin (ω t ) + C cos (ω t ) = A cos (ω t + φ )
dx
= ω B cos (ω t ) − ω C sin (ω t )
dt
d2x
2
2
2
=
−
ω
Bsin
ω
t
−
ω
C
cos
ω
t
=
−
ω
x ok
(
)
(
)
2
dt
12/9/09
6
x = A cos (ω t + ϕ )
 
But wait a minute...what does angular frequency ω have to
do with moving back & forth in a straight line ??
y
1
1
2
3
x
4
6
5
0
-1
1
2
π
3
π
2
4
θ
6
5
The spring’s motion with amplitude A is identical to the
x-component of a particle in uniform circular motion with
radius A.
12/9/09
7
 
A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t = 0
its speed is maximum, and is v = +2 m/s.
  What is the angular frequency of oscillation ω?
  What is the spring constant k?
vMAX 2 m s
ω=
=
= 20 s −1
A
10 cm
vMAX = ωA
Also: ω =
k
m
k = mω2
So k = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m
k
m
x
12/9/09
8
Use “initial conditions” to determine phase φ!
Suppose we are told x(0) = 0 , and x is
initially increasing (i.e. v(0) = positive):
x(t) = A cos(ωt + φ)
v(t) = -ωA sin(ωt + φ)
a(t) = -ω2A cos(ωt + φ)
φ = π/2 or -π/2
φ<0
x(0) = 0 = A cos(φ)
v(0) > 0 = -ωA sin(φ)
So φ = -π/2
π
k
12/9/09
cosθ sinθ
m
0
θ
2π
x
9
So we find φ= -π/2
x(t) = A cos(ωt - π/2 )
v(t) = -ωA sin(ωt - π/2 )
a(t) = -ω2A cos(ωt - π/2 )
x(t) = A sin(ωt)
v(t) = ωA cos(ωt)
a(t) = -ω2A sin(ωt)
A
π
k
m
0
12/9/09
x(t)
ωt
2π
-A
x
10
 
We already know that for a vertical spring
if y is measured from
the equilibrium position
 
The force of the spring is the negative
derivative of this function:
 
So this will be just like the horizontal case:
1 2
U = ky
2
k
d2y
−k y = ma = m 2
dt
Which has solution y = A cos(ωt + φ)
where
12/9/09
ω=
y=0
m
F = -ky
k
m
11
 
Recall that the torque due to gravity about the rotation (z) axis is
τ = -mgd
d = Lsin θ ≈ Lθ for small θ
so τ = -mg Lθ
But
d 2θ
τ=I 2
dt
z
τ = Iα, where I = mL2
θ
d 2θ
2
=
−
ω
θ
2
dt
where
ω=
g
L
Differential equation for simple harmonic motion!
θ = θ 0 cos (ω t + ϕ )
12/9/09
L
m
d
mg
12
 
 
You are sitting on a swing. A friend gives you a small push
and you start swinging back & forth with period T1.
Suppose you were standing on the swing rather than sitting.
When given a small push you start swinging back & forth
with period T2.
  Which of the following is true:
(a) T1 = T2
(b) T1 > T2
(c) T1 < T2
12/9/09
13
 
We have shown that for a simple pendulum
ω=
g
L
Recall the pendulum depended on the torque from gravity. Now the
moment arm is from the center of mass and the oscillation point so
L is shorter
2π
Since T =
ω
 
12/9/09
L
T = 2π
g
If we make a pendulum shorter, it oscillates faster (shorter period)
14
k
ω=
2
Force:
d s
2
= −ω s
2
dt
s
k
m
0
m
k
m
s
0
Solution:
s = A cos(ωt + φ)
12/10/09
ω=
g
L
s
L
15
x = A cos (ω t )
t
U
1 2 1 2
U = kx = kA cos 2 (ωt )
2
2
t
K
t
12/10/09
dx
v=
= −ω Asin (ω t )
dt
1
1
K = mv 2 = mω 2 A 2 sin 2 (ω t )
2
2
E = K +U = K Max =
1
1
mω 2 A 2 = U Max = k 2 A 2
2
2
16
1 2
U = kx
2
1 2
U = kx
2
1 2
K = mv
2
E =U + K
1 2 1 2
= kA = mv max
2
2
12/10/09
17
 
For both the spring and the pendulum, we can derive the SHM
solution by using energy conservation.
 
The total energy (K + U) of a
system undergoing SHM will
always be constant!
 
This is not surprising since there
are only conservative forces
present, hence K+U energy is conserved.
U
E
-A
12/10/09
0
K
U
A s
18
m
x
x=0
Etotal = 1/2 Mv2 + 1/2 kx2 = constant
KE
PE
KEmax = Mv2max /2 = Mω2A2 /2 =kA2 /2
PEmax = kA2 /2
Etotal = kA2 /2
12/10/09
19
1 2
U ≈ kx
2
for small x
12/10/09
20
 
 
Suppose we have some arbitrarily shaped
solid of mass M hung on a fixed axis, and
that we know where the CM is located and
what the moment of inertia I about the axis is.
The torque about the rotation (z) axis for
small θ is (using sin θ ≈ θ)
d 2θ
− MgRθ = I
τ = -Mgd ≈ -MgRθ
dt 2
τ
α
d 2θ
MgR
2
=
−
ω
θ
ω
=
where
I
dt 2
θ = θ 0 cos (ω t + ϕ )
12/10/09
z-axis
R
θ
xCM
d
Mg
21
 
 
Consider an object suspended by a wire attached
at its CM. The wire defines the rotation axis, and
the moment of inertia I about this axis is known.
The wire acts like a “rotational spring.”
  When the object is rotated, the wire is twisted.
This produces a torque that opposes the
rotation.
  In analogy with a spring, the torque produced
is proportional to the displacement: τ = -kθ
wire
θ
τ
I
12/10/09
22
 
Since τ = -kθ, τ = Iα becomes
d 2θ
−kθ = I 2
dt
d 2θ
2
=
−
ω
θ
2
dt
wire
θ
τ
where
ω=
k
I
I
This is similar to the “mass on spring” except
I has taken the place of m (no surprise).
12/10/09
23
 
 
 
 
12/10/09
In many real systems, nonconservative forces are present
  The system is no longer ideal
  Friction is a common nonconservative force
In this case, the mechanical energy of the system diminishes in
time, the motion is said to be damped
The amplitude decreases with time
The blue dashed lines on the graph represent the envelope of the
motion
24
 
 
 
 
One example of damped motion occurs
when an object is attached to a spring
and submerged in a viscous liquid
The retarding force can be expressed
as R = - b v where b is a constant
  b is called the damping
coefficient
The restoring force is – kx
From Newton’s Second Law
ΣFx = -k x – bvx = max
or,
d2x
dx
m 2 + b + kx = 0
dt
dt
12/10/09
25
d2x
dx
m 2 + b + kx = 0
dt
dt
When b is small enough, the solution to
this equation is:
x = A0 e−(b /2 m )t cos(ω ′t + δ )
with
2
⎛ b ⎞
ω′ = ω0 1 − ⎜
, ω0 =
⎟
⎝ 2mω 0 ⎠
k
m
(This solution applies when
b<2mω0.)
12/10/09
26
 
The position can be described by
x = A0 e
 
 
12/10/09
cos(ω ′t + δ )
The angular frequency is
ω′ =
 
−(b /2 m )t
k ⎛ b ⎞
−⎜
⎟
m ⎝ 2m ⎠
2
When the retarding force is small, the oscillatory character of the motion is
preserved, but the amplitude decreases exponentially with time
The motion ultimately ceases
27
 
The energy of a weakly damped harmonic oscillator decays
exponentially in time.
The potential energy is ½kx2:
(
x = A0 e
2
−(b /2 m )t
2 −(b / m )t
0
=Ae
cos(ω ′t + δ )
)
2
cos (ω ′t + δ )
2
Average over a period when (b/m)<<ω’:
Average of cos2 = ½, so averaged over a period,
x
12/10/09
2
2 −(b / m )t
0
≈A e
.
28
 
The kinetic energy is ½mv2:
⎛d
⎞
−(b /2 m )t
v =⎜
A0 e
cos(ω ′t + δ ) ⎟
⎝ dt
⎠
(
2
(
= −(b / 2m)A0 e
)
−(b /2 m )t
2
cos(ω ′t + δ ) − A0 e
−(b /2 m )t
sin(ω ′t + δ )
)
2
Average energy over a period when (b/m)<<ω’:
Average of cos2 = average of sin2 = ½, and average of
sinx cosx=0, so averaged over a period,
v
12/10/09
2
∝e
−(b / m )t
.
29
 
Both kinetic and potential energy (averaged over a period)
decay as e-(b/m)t.
So total energy is proportional to e-(b/m)t.
12/10/09
30
 
ω0 is also called the natural frequency of the system
If Rmax = bvmax < kA, the system is said to be underdamped
When b reaches a critical value bc such that bc / 2 m = ω0 , the system will not
oscillate
  The system is said to be critically damped
If Rmax = bvmax > kA and b/2m > ω0, the system is said to be overdamped
 
For critically damped and overdamped there is no angular frequency
 
 
 
12/10/09
31
 
 
 
 
It is possible to compensate for the loss of energy in a damped
system by applying an external sinusoidal force (with angular
frequency ω)
The amplitude of the motion remains constant if the energy input
per cycle exactly equals the decrease in mechanical energy in
each cycle that results from resistive forces
After a driving force on an initially stationary object begins to act,
the amplitude of the oscillation will increase
After a sufficiently long period of time, Edriving = Elost to internal
  Then a steady-state condition is reached
  The oscillations will proceed with constant amplitude
 
The amplitude of a driven oscillation is
  ω0 is the natural frequency of the undamped oscillator
12/10/09
32
 
 
 
 
 
 
 
 
When ω » ω0 an increase in amplitude occurs
This dramatic increase in the amplitude is
called resonance
The natural frequency w0 is also called the
resonance frequency
At resonance, the applied force is in phase
with the velocity and the power transferred to
the oscillator is a maximum
  The applied force and v are both
proportional to sin (ωt + φ)
  The power delivered is F . v
»  This is a maximum when F and v are
in phase
Resonance (maximum peak) occurs when driving
frequency equals the natural frequency
The amplitude increases with decreased damping
The curve broadens as the damping increases
The shape of the resonance curve depends on b
12/10/09
33
The amplitude of a system moving with simple harmonic
motion is doubled. The total energy will then be 12/10/09
4
times larger 2 times larger
the
half as much quarter as much
same as it was
1 2 1 2
kx + mv
2
2
at x = A, v = 0
1 2
U = kA
2
U=
34
A glider of mass m is attached to springs on both ends, which
are attached to the ends of a frictionless track. The glider
moved by 0.2 m to the right, and let go to oscillate. If m =
2 kg, and spring constants are k1 = 800 N/m and k2 = 500
N/m, the frequency of oscillation (in Hz) is approximately 12/10/09
6
Hz 2 Hz
4
8 Hz 10 Hz
Hz
ω
f =
=
2π
k/m
=
2π
Physics 201, Fall 2006, UW-Madison
800 + 500
2
= 4 Hz
2π
35
A 810-g block oscillates on the end of a spring whose force
constant is 60 N/m. The mass moves in a fluid which offers
a resistive force proportional to its speed – the
Fs + FR = ma
proportionality constant is 0.162 N.s/m.
  Write the equation of motion and solution.
  What is the period of motion?
0.730 s
d2x
dx
m 2 = −kx − b
dt
dt
Solution: x(t) = Ae
ω=
k ⎛ b ⎞
−⎜
⎟
m ⎝ 2m ⎠
  What is the fractional decrease in amplitude per cycle?
−
b
t
2m
cos(ω t + φ )
2
0.070
  Write the displacement as a function of time if at
t = 0, x = 0 and at t = 1 s, x = 0.120 m.
Solution: x(t) = 0.182e−0.100t sin(8.61t + φ )
12/10/09
Physics 201, Fall 2006, UW-Madison
36

Hooke`s Law Simple Harmonic Motion

Transcription

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