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
v x (t ) + x 2 (t ) = A2
(true at any time t)
k
The maximum speed vmax occurs when x = 0.
General Solution!
R. Field 11/7/2013
University of Florida
vmax =
k
A
m
x(t ) = A cos(ωt + φ )
v x (t ) = −ωA sin(ωt + φ )
k
ω =
2
a x (t ) = −ω A cos(ωt + φ ) spring m
PHY 2053
The phase angle φ
determines where the mass
m is at t = 0, x(t=0) = Acosφ.
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
R. Field 11/7/2013
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=
R. Field 11/7/2013
University of Florida
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
k
C =
spring m
ω= C
vmax = C A
2π
amax = CA 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
R. Field 11/7/2013
University of Florida
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 )
R. Field 11/7/2013
University of Florida
θ max = A
ξ= C
ωmax = C A
2π
=
T
α max = CA
1
f =
ω
=
2
π
f
C
T
PHY 2053
Page 5
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:
τ = Iα = −mgL sin θ
I = mL
α (t ) = −Cθ (t )
2
mgL
g
α (t ) = − 2 sin θ → − θ (t )
θ <<1 L
mL
SHM with period T given by T =
L
2π
= 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:
mgd cm
mgd cm
−
θ (t )
τ = Iα = −d cm mg sin θ α (t ) = − I sin θ θ→
<<1
I
mgd cm
α (t ) = −Cθ (t ) C = ω 2 =
I
I
2π
T
=
=
2
π
SHM with period T given by
C
R. Field 11/7/2013
University of Florida
PHY 2053
mgd cm
(physical pendulum)
Page 6
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
A2 =
2
m 2
v x (t ) + x 2 (t )
k
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
L
T = 2π
g
R. Field 11/7/2013
University of Florida
Tnew = 2π
L
L
= 2 × 2π
= 2T
( g / 4)
g
PHY 2053
Page 7
SHM: Example Problems
• 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? Answer: 0.5
f s = ma x ≤ μ s mg
μs =
R. Field 11/7/2013
University of Florida
amax = μ s g
2
amax = ω spring
A=
k
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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