oscillations chapter 15. simple harmonic motion (shm) systems
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OSCILLATIONSChapter 15
Simple Harmonic Motion (SHM) Systems
Period (T)repeat time of the motion
frequency ( f )rate of the repetition
Unit = Hz = 1 cycle/s = 1/s
f = 1/T
Amplitude (A)the maximum displacement from the midpoint
x = A cos(ωt + ϕ) ω = angular frequencyA measure of how rapidly the oscillations are occuring
T = 2π/ω
f = ω /2π
ϕ = phase constantOr initial phase angle
cos( ) sin( )dx d
v A t A tdt dt
22
2sin( ) cos( )
d x dA t A t
dtdt
x = A cos(ωt + ϕ)
22
2
d xx
dt
Maximum velocity
maxk
v A Am
2max
ka A A
m
Here’s a way of thinking about simple harmonic motion that will be very useful later. Suppose a particle is in uniform circular motion on a circle of radius A and with angular speed T. The projection of that motion onto any axis is a point moving in simple harmonic motion with amplitude A and circular frequency T.
A particle oscillates with simple harmonic motion, so that its displacement varies according to the expression x = (5 cm)cos(2t + π/6) where x is in centimeters and t is in seconds. At t = 0 find(a) the displacement of the particle,(b) its velocity, and(c) its acceleration.(d) Find the period and amplitude of the motion.
( ) cos( )x t A t
@ 0; (0) (5 )cos( ) 4.336
t x cm cm
2 (0) (5 )sin( ) 5
6
cmv cm
s s
2
2
2 (0) (5 )cos( ) 17.3
6
cma cm
s s
2 T s
5A cm
A 200-g block connected to a light spring for which the force constant is 5.00M/m is free to oscillate on a horizontal frictionless surface. The block is displaced 5.00 cm from equilibrium and released from rest.
k
m -3
5.00 /5.00 /
200x10
N mrad s
2T
21.26
5.00 /T s
rad s
Determine the maximum speed.
maxv A 2max (5.00 / )(5.00x10 ) 0.250 /v rad s m m s
What is the maximum acceleration?2
maxa A 2 2 2max (5.00 / ) (5.00x10 ) 1.25 /a rad s m m s
Express the position, velocity, and acceleration as functions of time.
sin( )v A t (0.250 / )sin(5.00 )v m s t
cos( )x A t (0.050 )cos(5.00 )x m t
2 sin( )a A t 2(1.25 / )cos(5.00 )a m s t
222 21 1 1 1
02 2 2 2
2 21 10 02 2
2 2 22 20 0
2
The total energy of a mass-spring system is
total energy cos( )
sin( ) cos( )
sin ( ) cos ( )2 2
Now, for a spring, so that
t
dxmv kx m k x t
dt
m x t k x t
m x k xt t
k
m
2 2 2 22 20 0otal energy = sin ( ) cos ( )
2 2
m x m xt t
The total energy of a simple harmonic oscillator is constant in time and proportional to the square of the amplitude
kinetic and potential energies (as fraction of total energy) for x = x0 cos(Tt)
U / (K+U), K/(K+U)
The frequency of a resonator is determined by the way energy shifts back and forth between kinetic and potential terms.
2 21 1
2 2
2
2
2 2
2 2
0
since . Using gives
0 0
E K U mv k x
dE dv dx dvmv k x v m k x
dt dt dt dt
dx dv d xv
dt dt dt
d x d x km k x x
dt dt m
These are called sinusoidal oscillations, or simple harmonic motion.
T h e so lu tio n to th e eq u atio n is
w h ere an d are
co n stan ts th a t d ep en d o n th e p o sitio n an d v e lo c ity
o f th e m ass a t tim e t = 0 .
2
0
d x
d t
k
mx
x t xk
mt x
2
0
0
( ) co s
For a mass-and-spring system, the motion is simple harmonic at an angular frequency of
The motion is always simple harmonic whenever the restoring force is a linear function of the oscillation parameter. Another way to say this is that the motion is simple harmonic whenever the potential energy is a quadratic function of the oscillation parameter.
km
For example, a pendulum has a kinetic energy of K = ½ mv2 and a potential energy of U = mgh.
x L vdx
d tL
d
d t
y L vdx
d tL
d
d t
K m v m v v
m L Ld
dtm L
d
dt
x
x
x y
s in co s
co s sin
co s sin
12
2 12
2 2
12
2 2 2 2
2
12
2
2
U m gh m g L
U m g L
1
1 12
2
12
2
co s
, co s
fo r sm all an g les
so th a t
K m Ld
dtU m g L
1
22
2
12
2,
K = ½ m L2 (dθ/dt)2 , U = ½ mgL θ2
Compare to
K = ½ m (dx/dt)2, U = ½ kx2
The circular frequency of the mass-spring system is
(½ k / ½ m)1/2 = ( k / m)1/2
By analogy, the circular frequency of the pendulum is
[½ mgL / ½ m L2]1/2 = [g / L]1/2