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