Physics 2211: Lecture 33 Todays Agenda

1
Physics 2211 Lecture 33Todays Agenda

• Introduction to Simple Harmonic Motion
• Horizontal spring mass
• The meaning of all these sines and cosines
• Vertical spring mass
• Initial conditions

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).
• This oscillation is called Simple Harmonic
  Motion, and is actually very easy to understand...

3
SHM Dynamics

• At any given instant we know that F ma must be
  true.
• But in this case F -kx and
  ma
• So -kx ma

a differential equation for x(t)!
4
SHM Dynamics...
Where w is the angular frequency of motion
Define
Guess a solution x A cos(?t f ) where f is a
constant (called the phase angle)
Check by plugging in solution to differential
equation
This works, so it must be a solution!
5
SHM Dynamics...
Shadow

• 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
2
3
3
?
0
x
4
-1
4
6
6
5
5
6
SHM Solution...

• Drawing of A cos(?t )
• A amplitude of oscillation,
• q (the stuff inside the parentheses) the
  phase

T 2?/?
A
?
??
?
??
-??
A
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SHM Solution...

• Drawing of A cos(?t ?)

?
?
??
?
??
-??
8
SHM Solution...

• Drawing of A cos(?t - ?/2)

? ??/2
A
?
??
?
??
-??
A sin(?t)!
9
Vertical Springs
(a)
(b)

• A spring is hung vertically. Its relaxed
  position is at y 0 (a). When a mass m is hung
  from its end, the new equilibrium position is ye
  (b).

j
k

• Recall that the force of a spring is Fs -kx.
  In case (b) Fs mg and x ye-kye - mg
  0 (ye lt 0)

y 0
y ye
-kye
mg
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Vertical Springs
(a)
(b)

• The potential energy of the spring-mass system is

j
k
y 0
y ye
choose C to make U0 at y ye
m
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Vertical Springs
(a)
(b)

• So

j
k
y 0
which can be written
y ye
m
12
Vertical Springs
(a)
(b)
j
k

• So if we define a new y? coordinate system such
  that y? 0 is at the equilibrium position, ( y?
  y - ye ) then we get the simple result

y? 0
m
?
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Vertical Springs
(a)
(b)

• If we choose y 0 to be at the equilibrium
  position of the mass hanging on the spring, we
  can define the potential in the simple form.
• Notice that g does not appear in this
  expression!!
• By choosing our coordinates and constants
  cleverly, we can hide the effects of gravity.

j
k
y 0
m
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SHO and Vertical Springs
Vertical Spring

• 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-ky ma

j
k
y 0
F -ky
Which has solution y A cos(?t ?)
where
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SHM So Far

• The most general solution is x A cos(?t ?)
• where A amplitude
• ? frequency
• ? phase angle
• For a mass on a spring
• 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!

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Velocity and Acceleration
Position x(t) A cos(?t ?) Velocity v(t)
-?A sin(?t ?) Acceleration a(t) -?2A
cos(?t ?)
by taking derivatives, since
xMAX A vMAX ?A aMAX ?2A
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Lecture 33, Act 1Simple Harmonic Motion

• A mass oscillates up down on a spring. Its
  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)
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Lecture 33, Act 1 Solution

• The slope of y(t) tells us the sign of the
  velocity since

• y(t) and a(t) have the opposite sign since a(t)
  -w2 y(t)

a lt 0v gt 0
a lt 0v lt 0
y(t)
(a)
(c)
t
(b)
a gt 0v gt 0
The answer is (c).
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Example

• 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 ??
• What is the spring constant k?

vMAX ?A
?
Also
k m?2
So k (2 kg) x (20 s -1) 2 800 kg/s2 800 N/m
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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(0) 0 A cos(?) ? ?/2 or -?/2 v(0) gt 0
-?A sin(?) ? lt 0
? -?/2
So
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Initial Conditions...
So we find ? -?/2!!
x(t) A cos(?t - ?/2 ) v(t) -?A sin(?t - ?/2
) a(t) -?2A cos(?t - ?/2 )
x(t) A sin(?t) v(t) ?A cos(?t) a(t) -?2A
sin(?t)
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Lecture 33, Act 2Initial Conditions

• A mass hanging from a vertical spring is lifted a
  distance d above equilibrium and released at t
  0. Which of the following describes its velocity
  and acceleration as a function of time?

(a) v(t) -vmax sin(wt) a(t) -amax
cos(wt)
k
y
(b) v(t) vmax sin(wt) a(t) amax
cos(wt)
d
t 0
(c) v(t) vmax cos(wt) a(t) -amax
cos(wt)
0
(both vmax and amax are positive numbers)
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Lecture 33, Act 2 Solution
Since we start with the maximum
possibledisplacement at t 0 we know that y
d cos(wt)
k
y
d
t 0
0
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Problem Vertical Spring

• A mass m 102 g is hung from a vertical spring.
  The equilibrium position is at y 0. The mass
  is then pulled down a distance d 10 cm from
  equilibrium and released at t 0. The measured
  period of oscillation is T 0.8 s.
• What is the spring constant k?
• Write down the equations for the position,
  velocity, and acceleration of the mass as
  functions of time.
• What is the maximum velocity?
• What is the maximum acceleration?

k
y
0
-d
m
t 0
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Problem Vertical Spring...

• What is k ?

k
y
0
-d
m
t 0
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Problem Vertical Spring...

• What are the equations of motion?
• At t 0,
• y -d -ymax
• v 0
• So we conclude

y(t) -d cos(?t) v(t) ?d sin(?t) a(t) ?2d
cos(?t)
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Problem Vertical Spring...
y(t) -d cos(?t) v(t) ?d sin(?t) a(t) ?2d
cos(?t)
?t
0
?
??
k
y
xmax d .1m vmax ?d (7.85 s-1)(.1m)
0.78 m/s amax ?2d (7.85 s-1)2(.1m) 6.2
m/s2
0
-d
m
t 0
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Recap of todays lecture

• Introduction to Simple Harmonic Motion
• Horizontal spring mass
• The meaning of all these sines and cosines
• Vertical spring mass
• Initial conditions