Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 31
Today’s Agenda

Today’s Topics
Simple Harmonic Motion – masses on springs
(Ch 15.1-2)
Energy of the SHO – Ch. 15.3
Pendula.
Physics 151: Lecture 31, Pg 1
See text: 15.1-2

Simple Harmonic Motion (SHM)
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).
k

This oscillation is called
Simple Harmonic Motion,
and is actually very easy
to understand...
k
k
m
m
m
Physics 151: Lecture 31, Pg 2
See text: 15.1-2
SHM Dynamics



At any given instant we know
that F = ma must be true.
F = -kx
k
a
But in this case F = -kx
m
d 2x
and ma = m
dt 2
2
So: -kx = ma = m d x
dt 2
x
d 2x
k
 x
2
m
dt
a differential equation for x(t) !
Physics 151: Lecture 31, Pg 3
See text: 15.1-2
SHM Dynamics...
d2x
k


x
2
m
dt

define
k
m
d2x
2



x
2
dt
Try the solution x = Acos(t)
dx
 A sint 
dt
d2x
2
2





A
cos

t



x
2
dt
this works, so it must be a solution !
Physics 151: Lecture 31, Pg 4
See text: 15.1-2
SHM Dynamics...

But wait a minute...what does angular frequency  have to
do with moving back & forth in a straight line ??
y = R cos  = R cos (t)
y
1
1
1
2
3
x
4
6
5
0
-1
2

3

2
4

6
5
Physics 151: Lecture 31, Pg 5
See text: 15.1-2
SHM Solution
d2x
2



x (which came from F=ma)
 We just showed that
2
dt
has the solution x = Acos(t) .

This is not a unique solution, though. x = Asin(t) is also a
solution.

The most general solution is a linear combination of these two
solutions!
x = Bsin(t)+ Ccos(t)

This is equivalent to: x = A cos(t+) where  is called a phase
Physics 151: Lecture 31, Pg 6
See text: 15.1-2
SHM Solution...


Drawing of Acos(t )
A = amplitude of oscillation
T = 2/
A

A



Physics 151: Lecture 31, Pg 7
See text: 15.1-2
SHM Solution...

Drawing of Acos(t + )





Physics 151: Lecture 31, Pg 8
See text: 15.1-2
SHM Solution...

Drawing of Acos(t - /2)

A




= Asin(t) !
Physics 151: Lecture 31, Pg 9
See text: 15.2
Velocity and Acceleration
x(t) = Acos(t + )
v(t) = -Asin(t + )
a(t) = -2Acos(t + )
Position:
Velocity:
Acceleration:
xMAX = A
vMAX = A
aMAX = 2A
k
by taking
derivatives,
since:
dx ( t )
v( t ) 
dt
a( t ) 
dv ( t )
dt
m
0
x
Physics 151: Lecture 31, Pg 10
Lecture 31, Act 1
Simple Harmonic Motion

A mass oscillates up & down on a spring. It’s position as a
function of time is shown below. At which of the points shown
does the mass have positive velocity and negative acceleration ?
y(t)
(a)
(c)
t
(b)
Physics 151: Lecture 31, Pg 11
Lecture 31, Act 1
Solution

The slope of y(t) tells us the sign of the velocity since v 

y(t) and a(t) have the opposite sign since a(t) = -2 y(t)
a<0
v<0
y(t)
dy
dt
a<0
v>0
(a)
(c)
t
(b)
a>0
v>0
The answer is (c).
Physics 151: Lecture 31, Pg 12
See text: 15.2
Example

A mass m = 2kg on a spring oscillates with amplitude A =
10cm. At t=0 its speed is maximum, and is v = +2 m/s.
What is the angular frequency of oscillation  ?
What is the spring constant k ?
v MAX 2 m s

 20 s 1
=
A
10 cm
vMAX = A
Also:  
k
m
k = m2
So k = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m
k
m
x
Physics 151: Lecture 31, Pg 13
See text: 15.2
Initial Conditions
Use “initial conditions” to determine phase  !
Suppose we are told x(0) = 0 , and x is
initially increasing (i.e. v(0) = positive):
x(t) = Acos(t + )
v(t) = -Asin(t + )
a(t) = -2Acos(t + )
So  = -/2

k

cos sin
m
0

x
Physics 151: Lecture 31, Pg 14
Lecture 31, Act 2
Initial Conditions

A mass hanging from a vertical spring is lifted a distance d
above equilibrium and released at t = 0. Which of the
following describe its velocity and acceleration as a function
of time (upwards is positive y direction):
(a) v(t) = vmax sin(t)
(b) v(t) = vmax cos(t)
(c) v(t) = - vmax sin(t)
a(t) = amax cos(t)
k
a(t) = -amax cos(t)
t=0
a(t) = -amax cos(t)
y
m
d
0
(both vmax and amax are positive numbers)
Physics 151: Lecture 31, Pg 15
See text: 15.2
Energy of the Spring-Mass System
We know enough to discuss the mechanical energy of the oscillating
mass on a spring.
Remember,
x(t) = Acos(t + )
v(t) = -Asin(t + )
a(t) = -2Acos(t + )
Kinetic energy is always K = 1/2 mv2
K = 1/2 m (-Asin(t + ))2
We also know what the potential energy of a spring is,
U = 1/2 k x2
U = 1/2 k (Acos(t + ))2
Physics 151: Lecture 31, Pg 16
See text: 15.3
Energy of the Spring-Mass System
Add to get E = K + U
1/2 m (A)2sin2(t + ) + 1/2 k (Acos(t + ))2
Remember that
k
k

2 
m
m
so, E = 1/2 kA2 sin2(t + ) + 1/2 kA2 cos2(t + )
= 1/2 kA2 ( sin2(t + ) + cos2(t + ))
= 1/2 kA2
E = 1/2 kA2
U~cos2
K~sin2

 
Physics 151: Lecture 31, Pg 17
See text: 15.1 to 15.3
SHM So Far

The most general solution is x = Acos(t + )
where A = amplitude
 = frequency
 = phase constant

For a mass on a spring  

k
m
The frequency does not depend on the amplitude !!!
We will see that this is true of all simple harmonic
motion !
The oscillation occurs around the equilibrium point where
the force is zero!
Physics 151: Lecture 31, Pg 18
See text: 15.4
The Simple Pendulum

A pendulum is made by
suspending a mass m at the end
of a string of length L. Find the
frequency of oscillation for small
displacements.
z

L
m
mg
Physics 151: Lecture 31, Pg 19
Aside: sin  and cos  for small 

A Taylor expansion of sin  and cos  about  = 0 gives:
sin    
3 5
3!

5!
 ...
and
cos  1 
2 4
2!

4!
 ...
So for  <<1, sin    and cos  1
Physics 151: Lecture 31, Pg 20
See text: 15-5

The Simple Pendulum...
Recall that the torque due to gravity about the
rotation (z) axis is  = -mgd.
d = Lsin   L
so  = -mg L

for small 
But  = I I = mL2
z

L
d
 mgL  mL
dt 2
g
d 2
2
   where  
m
2
L
dt
d
Differential equation for simple harmonic motion !
mg
 = 0 cos(t + )
2
2
Physics 151: Lecture 31, Pg 21
Lecture 31, Act 3
Simple Harmonic Motion


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
Physics 151: Lecture 31, Pg 22
Lecture 31, Act 3
Solution
Standing up raises the CM of the swing, making it shorter !
L
L1
2
T1
Since L1 > L2 we see that T1 > T2 .
T2
(b)
Physics 151: Lecture 31, Pg 23
Lecture 31, Act 4
Simple Harmonic Motion

Two clocks with basic timekeeping mechanism consist of
1) a mass on a string and 2) a simple pendulum. Both
have a period of 1s on Earth. When taken to the moon
which one of the statements below is correct ?
a)
b)
c)
d)
e)
the periods of both is unchanged.
one of them has a period shorter than 1 s.
the pendulum has a period longer than 1 s.
the mass-spring system has a period longer than 1s.
both c) and d) are true.
Physics 151: Lecture 31, Pg 24
See text: 15.4
The Rod Pendulum

A pendulum is made by suspending a thin rod of length L
and mass M at one end. Find the frequency of
oscillation for small displacements.
z

xCM
L
mg
Physics 151: Lecture 31, Pg 25
See text: 15.4
The Rod Pendulum...

The torque about the rotation (z) axis is
 = -mgd = -mg{L/2}sin -mg{L/2}for small 

1
In this case I  mL2
3
z
d
I
L
1
2 d 
So  = Ibecomes  mg   mL
2
3
dt 2
L/2
2


xCM
d 2
2




dt 2
where

3g
2L
L
d
mg
Physics 151: Lecture 31, Pg 26
Lecture 31, Act 4
Period

What length do we make the simple pendulum so that it
has the same period as the rod pendulum?
LS
3
(a) LS  LR
2
(b)
LR
2
LS  LR
3
(c) LS  LR
Physics 151: Lecture 31, Pg 27
See text: 15-5
Lecture 31, Act 4
Solution
LS
g
S 
LS
b) S = R if LS 
R 
3g
2 LR
LR
2
LR
3
Physics 151: Lecture 31, Pg 28
Recap of today’s lecture

Simple Harmonic Motion,
Example, block on a spring
Energy of SHM
Pendula are just like block/spring
Physics 151: Lecture 31, Pg 29