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Lecture 18:
Elasticity and Oscillations I

Simple Harmonic Motion: Definition

Springs: Forces

Springs: Energy

Simple Harmonic Motion: Equations
Simple Harmonic Motion
 Vibrations
Vocal cords when singing/speaking
String/rubber band
 Simple
Harmonic Motion
Restoring force proportional to displacement
Springs: F = -kx
Springs: Forces

Hooke’s Law: The force exerted by a spring is
proportional to the distance the spring is stretched
or compressed from its relaxed position.
FX = -k x
Where x is the displacement from
the relaxed position and k is the
constant of proportionality.
relaxed position
(equilibrium position)
FX = 0
+x
x=0
+x
Springs: Forces

Negative displacement:
Positive force
Positive acceleration

Zero displacement (at equilibrium):
Zero force
Zero acceleration

Positive displacement:
Negative force
Negative acceleration
Springs: Energy
 Force
of spring is Conservative:
F = -k x
W = ½ k x2
Work done only depends on initial and final position.
Potential Energy: Uspring = ½ k x2
U
A
mass oscillating on a spring:
Energy = U + K = constant
½ k x2 + ½ m v2 = total energy
0
m
x=0
x
x
+x
Springs: Energy

Maximum Negative displacement:
Maximum potential energy
Zero kinetic energy

Zero displacement (at equilibrium):
Zero potential energy
Maximum kinetic energy

Maximum Positive displacement:
Maximum potential energy
Zero kinetic energy
Simple Harmonic Motion
x(t) = [A]cos(t)
v(t) = -[A]sin(t)
a(t) = -[A2]cos(t)
xmax = A
vmax = A
amax = A2
x(t) = [A]sin(t)
OR
v(t) = [A]cos(t)
a(t) = -[A2]sin(t)
Period = T (seconds per cycle)
Frequency = f = 1/T (cycles per second)
Angular frequency =  = 2f = 2/T
For spring: 2 = k/m
Spring Example

A mass of m = 4 kg is attached to a spring with k = 16 N/m and
oscillates with an amplitude A = 0.15 m. What is the magnitude
of the acceleration of the mass when the potential energy stored in
the spring is equal to the kinetic energy of the mass?
First, use energy to find x where U = K:
» U + K = constant
Second, use Hooke’s Law to find force on the mass:
» F = -k x
Third, use Newton’s Second Law to find acceleration of mass:
» F=ma
Spring Example

A mass of m = 4 kg is attached to a spring with k = 16 N/m and
oscillates with an amplitude A = 0.15 m. What is the magnitude of
the acceleration of the mass when the potential energy stored in
the spring is equal to the kinetic energy of the mass?
First, use energy to find x where U = K:
»
»
»
»
»
U + K = constant
U + K = ½ k A2
U + U = ½ k A2
k x2 = ½ k A 2
x2 = ½ A2
1 2
x
A
2
(since total energy is ½ k A2)
(since K = U when they are equal)
(since U = ½ k x2 when they are equal)
(simplify)
= 0.106 m
Spring Example

A mass of m = 4 kg is attached to a spring with k = 16 N/m and
oscillates with an amplitude A = 0.15 m. What is the magnitude of
the acceleration of the mass when the potential energy stored in
the spring is equal to the kinetic energy of the mass?
Second, use Hooke’s Law to find force on the mass:
» F=kx
(we are only concerned with magnitude)
1 2
F k
A
2
= 1.70 N
Spring Example

A mass of m = 4 kg is attached to a spring with k = 16 N/m and
oscillates with an amplitude A = 0.15 m. What is the magnitude of
the acceleration of the mass when the potential energy stored in
the spring is equal to the kinetic energy of the mass?
Third, use Newton’s Second Law to find acceleration of mass:
» F=ma
(we are only concerned with magnitude)
k 1 2
a
A = 0.424 m/s2
m 2
Young’s Modulus

Spring
F=kx
What happens to “k” if cut spring in half?
1) ½
2) same
3) 2
k is inversely proportional to length!
 Define

Strain = DL / L
Stress = F/A

Now
Stress = Y Strain
F/A = Y DL/L
k = Y A/L from F = k x

Y (Young’s Modules) independent of L
15
What does moving in a circle have to do with
moving back & forth in a straight line ??
x = R cos  = R cos (t)
since  =  t
x
x
1
1
2
R
8
R
3
Movie

2
8
7
3
y
7
4
6
5
0
-R

2

4

3
2
6
5
46
(A)
(B)
(C)
(A)
(B)
(C)