Simple Harmonic Motion
Things that vibrate
§ 13.1–13.3
Hooke’s Law
• Force Law: F = –kx
• F = force exerted by the spring
• k = spring constant (characteristic of the
particular spring)
• x = distance spring is displaced from
equilibrium
Conditions of Motion
• Newton’s second law: F = ma
• Force F depends on position by Hooke’s
law: F = –kx
–kx = ma
• tells how motion changes at each position
d2x
–kx = m 2
dt
• second-order ordinary differential equation
Group Work
A Hooke’s law spring is fixed at one end and
has a mass attached to the other end. The
spring is initially at its neutral position. The
mass is pulled, stretching the spring. The
mass is then released.
a. Plot the mass’s position vs. time.
b. Select at least four events on the plot. At each
event, draw arrows to indicate the mass’s
velocity and acceleration.
CPS Question
The net force on a Hooke’s law object is
A. zero at the top and bottom
B. maximum at the top and bottom
C. minimum but not zero at the top and
bottom
CPS Question
The acceleration of a Hooke’s law object is
A. zero at the top and bottom
B. maximum at the top and bottom
C. minimum but not zero at the top and
bottom
CPS Question
The velocity of a Hooke’s law object is
A. maximum at the equilibrium position
B. maximum at the top and bottom
C. maximum midway between equilibrium
and top or bottom
Simulation Spreadsheet
From class web site, right-click on the “Numerical
simulation” link.
Enter parameters to produce 1–2 cycles.
• Describe how position changes with time.
• What is the velocity when position is at an extreme? At
0?
• What is the force when velocity is at an extreme? At 0?
• Where is max/min U?
• Where is max/min K?
Simulation Spreadsheet
From class web site, right-click on the “Numerical
simulation” link.
Enter parameters to produce 1–2 cycles.
• What happens to the period when k changes?
• What happens to the period when m changes?
CPS Question
All other things being equal, if the mass of a
Hooke’s law oscillator increases, its period
A. decreases.
B. does not change.
C. increases.
CPS Question
All other things being equal, if the stiffness k
of a Hooke’s law spring increases, its period
A. decreases.
B. does not change.
C. increases.
Group Work
• First MasteringPhysics problem, parts A-E.
Uniform Circular Motion
• Centripetal force F = mv2/r inwards
• Constant magnitude F0; direction depends
on position q
F0
F0
F0
F0
F0
F0
• Force in y-direction is proportional to –y
Uniform Circular Motion
• Angle changes at a steady rate.
• Projection on y-axis has Hooke’s law
force.
• So, projection on y-axis must have
Hooke’s law motion too!
• What is the projection of an angle on the
y-axis?
Group Work
Verify that
d2x
–kx = m 2
dt
if x(t) = A cos (wt + f),
where
• A and f are any real constants
• w = k/m
Equation of Motion
x(t) = A cos (wt + f),
• Amplitude A
• Angular frequency w
• Initial phase angle f
Period and Frequency
• Period T
– time of one cycle (units: s)
• Frequency f
– cycles per unit time (units: 1/s = Hz)
– f = 1/T
• Angular frequency w
– radians per unit time (units: 1/s = Hz)
– w = 2pf
– w = k/m
Another form
x(t) = A cos (wt + f) = C cos (wt) + S sin (wt)
where
C = A cos (f) and S = –A sin (f)
Group Work
• Fourth MasteringPhysics problem
(Exercise 13.4).
Group Work
• Second MasteringPhysics problem
(“Harmonic Oscillator Acceleration”).
Group Work
• Fifth MasteringPhysics problem (Exercise
13.10), parts A–C.
Initial Conditions
• Given m, k, x0, and v0, how does one find
the equations of motion?
– m and k give w.
– x0, v0, and w give A.
– x0/A and v0/(Aw) give f.
• Now you can do MasteringPhysics
Exercise 13.26.
• Can practice/explore with “SHM
parameters” spreadsheet
CPS Question
The potential energy of an oscillating mass
is greatest
A. at its extreme positions.
B. at its equilibrium (middle) position.
C. between the middle and an extreme
position.
CPS Question
The kinetic energy of an oscillating mass is
greatest
A. at its extreme positions.
B. at its equilibrium (middle) position.
C. between the middle and an extreme
position.
Constant Mechanical Energy
E = 1/2 kx2 + 1/2 mv2
Energy
• Potential energy of a stretched spring :
PE =
1
2
kx2
• Conservation of energy:
PE + KE = constant =
1
2
kA2
where A is the oscillation amplitude.
(This of course ignores the sullen reality of
energy dispersal by friction and drag.)
Energy
KE = 0
PE = max
KE = max
PE = 0
y
time
KE = 0
PE = max
Group Work
• First MasteringPhysics problem, parts F
and G
Effect of Gravity
• Less than you might expect:
• Changes equilibrium position x = 0
• Does not change k
Spring + Gravity
force
spring alone
gravity
0
0
position
Spring + Gravity
force
spring alone
gravity
0
spring + gravity
0
0
position
Spring + Gravity
different equilibrium position
net force
same k
0
0
position