Simple Harmonic Motion: SHM
• Linear Restoring Force: Ideal Spring
k
Fx = ma x = − kx
a x (t ) = − x(t )
m
k
k
Spring Constant k
amax = xmax = A
m
m
• Energy Conservation:
E = KE + U = mv + kx
1
2
2
x
1
2
At xmax = A vx = 0 hence
E = 12 kA2 = 12 mvx2 + 12 kx 2
Amplitude A
2
constant (independent of time)
m 2
vx (t ) + x 2 (t ) = A2
(true at any time t)
k
The maximum speed vmax occurs when x = 0.
General Solution!
x(t ) = A cos(ωt + φ )
v x (t ) = −ωA sin(ωt + φ )
a x (t ) = −ω A cos(ωt + φ )
2
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University of Florida
PHY 2053
vmax =
ω =
spring
k
A
m
k
m
The phase angle φ
determines where the mass
m is at t = 0, x(t=0) = Acosφ
φ.
If x(t=0) = A then φ = 0.
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Uniform Circular Motion & SHM
• Uniform Circular Motion: θ (t ) = ωt
A
θ (t)
x(t)
x-axis
Project uniform circular motion (constant angular velocity ω) of a
vector with length A onto the x-axis and you get SHM!
x(t ) = A cos[θ (t )] = A cos(ωt )
x(t) •
Amplitude
Simple Harmonic Motion (SHM):
If x(t=0) = A then
x(t ) = A cos(ωt )
v x (t ) = −ωA sin(ωt )
a x (t ) = −ω 2 A cos(ωt )
xmax = A
vmax = ωA
amax = ω 2 A
k
a x (t ) = −ω x(t ) = − x(t )
spring
m
2
ω =
spring
k
m
The period T is the time is takes for one circular revolution:
Time t
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University of Florida
T=
2π
ω
T = period (in s)
1
f =
T
f = frequency (in Hz)
PHY 2053
ω = 2πf
ω = angular frequency
(in rad/sec)
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SHM: Graphical Representation
If x(t=0) = A then
x(t ) = A cos(ωt )
v x (t ) = −vmax sin(ωt )
a x (t ) = −amax cos(ωt )
xmax = A
vmax = ωA
amax = ω 2 A
T=
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PHY 2053
2π
ω
1
f =
T
ω =
spring
k
m
ω = 2πf
Page 3
SHM: General Solution
If the acceleration ax(t) and the position x(t) are related as follows:
a x (t ) = −Cx(t )
where C is some constant then
x(t ) = A cos(ωt + φ )
v x (t ) = −vmax sin(ωt + φ )
a x (t ) = −amax cos(ωt + φ )
If x(t=0) = A then φ = 0:
xmax = A
vmax = C A
amax = CA
k
C =
spring m
ω= C
2π
T=
C
1
f =
ω = 2πf
T
If x(t=0) = 0 and vx(t=0) > 0 then φ = π/2:
x(t ) = A cos(ωt )
x(t ) = A sin(ωt )
v x (t ) = −vmax sin(ωt )
a x (t ) = −amax cos(ωt )
v x (t ) = vmax cos(ωt )
a x (t ) = − amax sin(ωt )
cos( A + B ) = cos A cos B + sin A sin B
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PHY 2053
Page 4
SHM: General Solution
• Angular Oscillations SHM:
If the angular acceleration α(t) and the angular position θ(t) are related as follows:
α (t ) = −Cθ (t )
where C is some constant then
θ (t ) = A cos(ξt + φ )
ω (t ) = −ωmax sin(ξt + φ )
α (t ) = −α max cos(ξt + φ )
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θ max = A
ξ= C
ωmax = C A
2π
α max = CA T =
1
f =
ω = 2πf
C
T
PHY 2053
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The Pendulum: Small Oscillations SHM
• Simple Pendulum:
axis
Small pendulum bob with mass m on string of lengh L and
negligible mass. Calculate the torque about the axis of rotation as
follows:
mgL
g
α (t ) = − 2 sin θ → − θ (t )
θ <<1 L
mL
τ = Iα = −mgL sin θ
I = mL
α (t ) = −Cθ (t )
2
SHM with period T given by T =
2π
L
= 2π
g
C
g
C=
L
θ (t) L
m
θ
mg
(simple pendulum)
• Physical 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:
τ = Iα = − d cm mg sin θ
α (t ) = −Cθ (t )
α (t ) = −
C = ω2 =
mgd cm
I
mgd cm
mgd cm
sin θ → −
θ (t )
θ
<<
1
I
I
SHM with period T given by T =
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2π
I
= 2π
mgd cm
C
PHY 2053
(physical pendulum)
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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
E = kA = mv + kx
1
2
A=
2
1
2
2
x
1
2
m 2
A = v x (t ) + x 2 (t )
k
2
2
m 2
(2kg )
v x (t ) + x 2 (t ) =
(40m / s ) 2 + (3m) 2 = 5m
k
(200 N / m)
• 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
T = 2π
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L
g
Tnew = 2π
L
L
= 2 × 2π
= 2T
( g / 4)
g
PHY 2053
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Exam 2 Fall 2011: Problem 52
• In the figure, 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? (Assume that the
system is near the surface of the Earth.)
ω spring =
Answer: 0.5
% Right: 26%
f s = ma x ≤ µ s FN
amax = µ s g
amax = ω
FN = mg
µs =
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2
spring
k
( M + m)
k
A=
A
( M + m)
k
196 N / m
A=
(0.5m) = 0.5
( M + m) g
(20kg )(9.8m / s )
PHY 2053
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Exam 3 Spring 2013: Problem 37
• A block of mass M = 4 kg is at rest on a horizontal
v
frictionless surface and is connected to an ideal
spring as shown in the figure. A 2-gram bullet
traveling horizontally at 290 m/s strikes the block and
becomes embedded in the block. If the period of the
subsequent simple harmonic motion of the bulletblock system is 1.73 s, what is the amplitude of the
oscillations (in cm)?
mv = (m + M )V
Answer: 4.0
% Right: 61%
1
2
V = mv /( m + M )
2 2
1
kA2 = 12 (m + M )V 2 + 12 kx 2 x→
m
v /(m + M )
2
=0
ω=
k
(m + M )
Ideal
Spring
M
V
Ideal
Spring
M
T=
2π
ω
k = (m + M )ω 2 = 4π 2 (m + M ) / T 2
m 2v 2
m 2 v 2T 2
mvT
(0.002kg )(290m / s )(1.73s )
A=
=
=
=
≈ 4.0cm
k (m + M )
4π 2 (m + M ) 2 2π (m + M )
2π (4.002kg )
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PHY 2053
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