Physics 2211, Spring 2002

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Physics 2211 Lecture 35Todays Agenda

• Recap of last lecture
• Using initial conditions to solve problems
• Problem Vertical Spring
• The general physical pendulum
• Energy in SHM

2
SHM and Springs
Solution
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Velocity and Acceleration
Position Velocity Acceleration
by taking derivatives, since
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ExampleSimple 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?

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Example 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 lt 0
a lt 0v gt 0
a gt 0v gt 0
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Example

• A mass m 2 kg on a spring oscillates with
  amplitude A 10 cm. At t 0 s its speed is
  maximum and is v 2 m/s.
• What is the angular frequency of oscillation ? ?
• What is the spring constant k ?

Also
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 and
amplitude A.
Suppose we are told x(0) 0 and x is initially
increasing (i.e., v(0) v0)
? ?/2 or -?/2 ? lt 0
? -?/2 A v0/w
So
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Initial Conditions
So we find ? -?/2 and A v0/w
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Example 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 describes its velocity
  and acceleration as a function of time?

(a) v(t) -vmax sin(w t) a(t) -amax cos(w
t)
(b) v(t) vmax sin(w t) a(t) amax cos(w
t)
(c) v(t) vmax cos(w t) a(t) -amax cos(w
t)
(both vmax and amax are positive numbers)
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Example Solution
Since we start with the maximum
possibledisplacement at t 0 we know that
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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 a
  function of time.
• What is the maximum velocity?
• What is the maximum acceleration?

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Problem Vertical Spring

• What is k ?

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Problem Vertical Spring

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

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Problem Vertical Spring
?t
0
?
??
k
y
ymax d 0.1 m vmax ?d (7.85 s-1)(0.1 m)
0.78 m/s amax ?2d (7.85 s-1)2(0.1 m) 6.2
m/s2
0
-d
m
t 0
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Review of Simple Pendulum

• Using ? I? and sin ? ? ? for small ?

?
?
I
We found
where
Which has SHM solution ? ?0 cos(? t ? )
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Review of Rod Pendulum

• Using ? I? and sin ? ? ? for small ?

We found
where
Which has SHM solution ? ?0 cos(? t ?)
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General Physical Pendulum

• 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
  ? is (sin? ? )
  
  

z-axis
R
?
x
CM
d
Mg
where
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Example Physical Pendulum

• A pendulum is made by hanging a thin hoola-hoop
  of diameter D on a small nail.
• What is the angular frequency of oscillation of
  the hoop for small displacements? (ICM mR2 for
  a hoop)

pivot (nail)
(a) (b) (c)
D
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Example Solution

• The angular frequency of oscillation of the hoop
  for small displacements will be given by

Use parallel axis theorem I Icm mR2
I mR2 mR2 2mR2
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Simple Harmonic Motion Summary
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Energy in SHM

• 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.

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Energy in SHM(example mass/spring)

• Energy is conserved (a constant of the motion).

• Energy is proportional to A2.

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Recap of todays lecture

• Recap of last lecture
• Using initial conditions to solve problems
• Problem Vertical Spring
• The general physical pendulum
• Energy in SHM
• Review Chapter 14 in Tipler