Transcript Simple Harmonic Motion Physics 202 Professor Vogel (Professor Carkner’s

Simple Harmonic Motion
Physics 202
Professor Vogel
(Professor Carkner’s
notes, ed)
Lecture 2
Simple Harmonic Motion
 Any motion that repeats itself in a sinusoidal fashion
 e.g. a mass on a spring
 A mass that moves between +xm and -xm
 with period T
 Properties vary from a positive maximum to a
negative minimum
 Position (x)
 Velocity (v)
 Acceleration (a)
 The system undergoing simple harmonic motion
(SHM) is a simple harmonic oscillator (SHO)
SHM Snapshots
Key Quantities
Frequency (f) -- number of complete
oscillations per unit time
Unit=hertz (Hz) = 1 oscillation per second = s-1
Period (T) -- time for one complete oscillation
T=1/f
Angular frequency (w) -- w = 2pf = 2p/T
Unit = radians per second (360 degrees = 2p
radians)
We use angular frequency because the
motion cycles
Equation of Motion
What is the position (x) of the mass at time (t)?
The displacement from the origin of a particle
undergoing simple harmonic motion is:
x(t) = xmcos(wt + f)
Amplitude (xm) -- the maximum displacement
from the center
Phase angle (f) -- offset due to not starting at
x=xm (“start” means t=0)
Remember that (wt+f) is in radians
SHM Formula Reference
SHM in Action
Consider SHM with f=0:
x = xmcos(wt)
Remember w=2p/T
t=0, wt=0, cos (0) = 1
x=xm
t=1/2T, wt=p, cos (p) = -1
x=-xm
t=T, wt=2p, cos (2p) = 1
x=xm
Phase
The phase of SHM is the quantity in
parentheses, i.e. cos(phase)
The difference in phase between 2 SHM
curves indicates how far out of phase
the motion is
The difference/2p is the offset as a
fraction of one period
Example: SHO’s f=p & f=0 are offset 1/2
period
They are phase shifted by 1/2 period
Amplitude, Period and Phase
Velocity
 If we differentiate the equation for
displacement w.r.t. time, we get velocity:
 v(t)=-wxmsin(wt + f)
 Why is velocity negative?
 Since the particle moves from +xm to -xm the
velocity must be negative (and then positive in the
other direction)
 Velocity is proportional to w
 High frequency (many cycles per second) means
larger velocity
Acceleration
If we differentiate the equation for
velocity w.r.t. time, we get acceleration
a(t)=-w2xmcos(wt + f)
This equation is similar to the equation
for displacement
Making a substitution yields:
a(t)=-w2x(t)
x, v and a
 Consider SMH with f=0:
 x = xmcos(wt)
 v = -wxmsin(wt) = -vmsin(wt)
 a = -w2xmcos(wt) = -amcos(wt)
 When displacement is greatest
(cos(wt)=1), velocity is zero and
acceleration is maximum
 Mass is momentarily at rest, but
being pulled hard in the other
direction
 When displacement is zero
(cos(wt)=0), velocity is
maximum and acceleration is
zero
 Mass coasts through the middle at
high speed
Force
Remember that: a=-w2x
But, F=ma so,
F=-mw2x
Since m and w are constant we can write the
expression for force as:
F=-kx
Where k=mw2 is the spring constant
This is Hooke’s Law
Simple harmonic motion is motion where
force is proportional to displacement but
opposite in sign
Why is the sign negative?
Linear Oscillator
A simple 1-dimensional SHM system is
called a linear oscillator
Example: a mass on a spring
In such a system, k=mw2
We can thus find the angular frequency
and the period as a function of m and k
k
ω
m
m
T2π
k
Linear Oscillator
Application of the Linear
Oscillator: Mass in Free Fall
A normal spring scale does not work in
the absence of gravity
However, for a linear oscillator the
mass depends only on the period and
the spring constant:
T=2p(m/k)0.5
m/k=(T/2p)2
m=T2k/4p2
SHM and Energy
A linear oscillator has a total energy E,
which is the sum of the potential and
kinetic energies (E=U+K)
U and K change as the mass oscillates
As one increases the other decreases
Energy must be conserved
SHM Energy Conservation
Potential Energy
Potential energy is the integral of force
UFdxkxdx 1kx 2
2
From our expression for x
U=½kxm2cos2(wt+f)
Kinetic Energy
Kinetic energy depends on the velocity,
K=½mv2 = ½mw2xm2 sin2(wt+f)
Since w2=k/m,
K = ½kxm2 sin2(wt+f)
The total energy E=U+K which will
give:
E= ½kxm2
Next Time
Read: 15.4-15.6