R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy...
-
Upload
leon-gardner -
Category
Documents
-
view
215 -
download
0
Transcript of R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy...
![Page 1: R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy Conservation: Linear Restoring Force: At x max = A v x = 0 hence.](https://reader036.fdocuments.in/reader036/viewer/2022082612/5697bf771a28abf838c818ad/html5/thumbnails/1.jpg)
R. Field 11/7/2013 University of Florida
PHY 2053 Page 1
Simple Harmonic Motion: SHM
)()( txm
ktax
• Energy Conservation:
• Linear Restoring Force:
kxmaF xx
At xmax = A vx = 0 hence
2212
21 kxmvUKEE x
Spring Constant k
2212
212
21 kxmvkAE x
constant (independent of time)
Am
kx
m
ka maxmax
222 )()( Atxtvk
mx
(true at any time t)
Amplitude A
The maximum speed vmax occurs when x = 0.
)cos()(
)sin()(
)cos()(
2
tAta
tAtv
tAtx
x
x
m
kspring
General Solution!
Am
kv max
The phase angle determines where the mass m is at t = 0, x(t=0) = Acos. If x(t=0) = A then = 0.
Ideal Spring
![Page 2: R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy Conservation: Linear Restoring Force: At x max = A v x = 0 hence.](https://reader036.fdocuments.in/reader036/viewer/2022082612/5697bf771a28abf838c818ad/html5/thumbnails/2.jpg)
R. Field 11/7/2013 University of Florida
PHY 2053 Page 2
Uniform Circular Motion & SHM
x-axis x(t)
(t)
A • Uniform Circular Motion:
Time t
x(t)
tt )(
Tf
1
)cos()](cos[)( tAtAtx
Project uniform circular motion (constant angular velocity ) of a vector with length A onto the x-axis and you get SHM!
2
T f 2
)cos()(
)sin()(
)cos()(
2 tAta
tAtv
tAtx
x
x
If x(t=0) = A then
The period T is the time is takes for one circular revolution:
• Simple Harmonic Motion (SHM):
)()()( 2 txm
ktxtaspring
x m
kspring
Aa
Av
Ax
2max
max
max
Amplitude
T = period (in s) f = frequency (in Hz) = angular frequency (in rad/sec)
![Page 3: R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy Conservation: Linear Restoring Force: At x max = A v x = 0 hence.](https://reader036.fdocuments.in/reader036/viewer/2022082612/5697bf771a28abf838c818ad/html5/thumbnails/3.jpg)
R. Field 11/7/2013 University of Florida
PHY 2053 Page 3
SHM: Graphical Representation
)cos()(
)sin()(
)cos()(
max
max
tata
tvtv
tAtx
x
x
If x(t=0) = A then
m
kspring
Aa
Av
Ax
2max
max
max
Tf
1
2
T f 2
![Page 4: R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy Conservation: Linear Restoring Force: At x max = A v x = 0 hence.](https://reader036.fdocuments.in/reader036/viewer/2022082612/5697bf771a28abf838c818ad/html5/thumbnails/4.jpg)
R. Field 11/7/2013 University of Florida
PHY 2053 Page 4
SHM: General Solution
)cos()(
)sin()(
)cos()(
max
max
tata
tvtv
tAtx
x
x
If the acceleration ax(t) and the position x(t) are related as follows:
C
CAa
ACv
Ax
max
max
max
Tf
1
CT
2 f 2
)()( tCxtax where C is some constant then
)sin()(
)cos()(
)sin()(
max
max
tata
tvtv
tAtx
x
x
BABABA sinsincoscos)cos(
)cos()(
)sin()(
)cos()(
max
max
tata
tvtv
tAtx
x
x
m
kCspring
If x(t=0) = A then = 0: If x(t=0) = 0 and vx(t=0) > 0 then = /2:
![Page 5: R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy Conservation: Linear Restoring Force: At x max = A v x = 0 hence.](https://reader036.fdocuments.in/reader036/viewer/2022082612/5697bf771a28abf838c818ad/html5/thumbnails/5.jpg)
R. Field 11/7/2013 University of Florida
PHY 2053 Page 5
SHM: General Solution
)cos()(
)sin()(
)cos()(
max
max
tt
tt
tAt
If the angular acceleration (t) and the angular position (t) are related as follows:
C
CA
AC
A
max
max
max
Tf
1
CT
2 f 2
)()( tCt where C is some constant then
• Angular Oscillations SHM:
![Page 6: R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy Conservation: Linear Restoring Force: At x max = A v x = 0 hence.](https://reader036.fdocuments.in/reader036/viewer/2022082612/5697bf771a28abf838c818ad/html5/thumbnails/6.jpg)
R. Field 11/7/2013 University of Florida
PHY 2053 Page 6
The Pendulum: Small Oscillations SHM
(t)
mg
L
m
axis • Simple Pendulum:Small pendulum bob with mass m on string of lengh L and negligible mass. Calculate the torque about the axis of rotation as follows:
2mLI sinmgLI
)()( tCt SHM with period T given by
g
L
CT
22
)()( tCt
• Physical Pendulum:
(simple pendulum)
Moment of inertia, I, Length L, mass m, distance from axis of rotation to the center-of-mass, dcm. Calculate the torque about the axis of rotation as follows:
I
mgdC cm 2
sinmgdI cm )(sin)(1
tI
mgd
I
mgdt cmcm
SHM with period T given bycmmgd
I
CT
22
(physical pendulum)
L
gC
)(sin)(12
tL
g
mL
mgLt
![Page 7: R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy Conservation: Linear Restoring Force: At x max = A v x = 0 hence.](https://reader036.fdocuments.in/reader036/viewer/2022082612/5697bf771a28abf838c818ad/html5/thumbnails/7.jpg)
R. Field 11/7/2013 University of Florida
PHY 2053 Page 7
SHM: Example Problems• A simple harmonic oscillator consists of a block of mass 2 kg attached
to a spring of spring constant 200 N/m. If the speed of the block is 40 m/s when the displacement from equilibrium is 3 m, what is the amplitude of the oscillations? Answer: 5m
• A simple pendulum has a length L. If its period is T when it is on the surface of the Earth (gravitational acceleration g ), what is its period when it is on the surface of a planet with gravitational acceleration equal to g/4? Answer: 2T
2212
212
21 kxmvkAE x )()( 222 txtv
k
mA x
mmsmmN
kgtxtv
k
mA x 5)3()/40(
)/200(
)2()()( 2222
g
LT 2 T
g
L
g
LTnew 222
)4/(2
![Page 8: R. Field 11/7/2013 University of Florida PHY 2053Page 1 Simple Harmonic Motion: SHM Energy Conservation: Linear Restoring Force: At x max = A v x = 0 hence.](https://reader036.fdocuments.in/reader036/viewer/2022082612/5697bf771a28abf838c818ad/html5/thumbnails/8.jpg)
R. Field 11/7/2013 University of Florida
PHY 2053 Page 8
SHM: Example Problems• Two blocks (m = 5 kg and M = 15 kg) and a
spring (k = 196 N/m) are arranged on a horizontal frictionless surface. If the smaller block begins to slip when the amplitude of the simple harmonic motion is greater than 0.5 m, what is the coefficient of static friction between the two blocks? Answer: 0.5
mgmaf sxs ga smaxA
mM
kAa spring )(
2max
5.0)5.0()/8.9)(20(
/196
)(
m
smkg
mNA
gmM
ks