Oscillatory Motion

Serway & Jewett (Chapter 15)

M Equilibrium position: no net force

M

The spring force is always directed back towards equilibrium (hence called the ‘restoring force’). This leads to an oscillation of the block about the equilibrium position.

M

For an ideal spring, the force is proportional to displacement. For this particular force behaviour, the oscillation is simple harmonic motion.

x

F = -kx

)cos( tAxSHM: x(t)

t

A

-A

T

A = amplitude

= phase constant

= angular frequency

A is the maximum value of x (x ranges from +A to -A).

gives the initial position at t=0: x(0) = A cos .

is related to the period T and the frequency f = 1/T

T (period) is the time for one complete cycle (seconds).Frequency f (cycles per second or hertz, Hz) is the number of complete cycles per unit time.

2 2

fT

units: radians/second or s-1

(“omega”) is called the angular frequency of the oscillation.

Velocity and Acceleration

2MAX

MAX

22

:Note

)cos()(

)sin()(

)cos( )(

Aa

Av

xtAdt

dvta

tAdt

dxtv

tAtx

a(t) 2 x(t)

Position, Velocity and Acceleration

x(t) t

v(t) t

a(t) t

Question: Where in the motion is the velocity largest? Where in the motion is acceleration largest?

Example:

SHM can produce very large accelerations if the frequency is high.

Engine piston at 4000 rpm, amplitude 5 cm:

m/s 8770

)s 419(m 05.0

sec 419

cradians/se 60/24000

cos)cm00.5(

2

212MAX

1

xa

tx

Simple Harmonic Motion

SHM: )cos( tAx

a(t) = 2 x(t)

dtdv

adtdx

v , We can differentiate x(t):

and find that acceleration is proportional to displacement:

But, how do we know something will obey x=Acos(t) ???

Mass and Spring

xmk

dtxd

a 2

2 M

x

F = -kxmakxF Newton’s 2nd Law:

so

This is a 2nd order differential equation for the function x(t). Recall

that for SHM, a 2 x : identical except for the proportionality

constant. So the motion of the mass will be SHM:

x(t) = A cos (t + ), and to make the equations for acceleration match, we require that

mk2 , or

mk (and = 2f, etc.).

Note: The frequency is independent of amplitude

Example: Elastic bands and a mass. A mass, m, is attached to two elastic bands. Each has tension T. The system is on a frictionless horizontal surface. Will this behave like a SHO?

Quiz

The ball oscillates vertically on a single spring with period T0 . If two identical springs are used, the new period will be

A) longerB) shorter

by a factor of

i) ii) 2iii) 4

2

Quiz

The ball oscillates vertically on a single spring with period T0 . If two identical springs are used, the period will be

A) longerB) shorter

by a factor of

i) ii) 2iii) 4iv) 1

2

Quiz

The amplitude of the oscillation gradually increases till block B starts to slip. At what A does this happen?(there is no friction between the large block and the surface)

a) Any Ab) 5 cmc) 50 cmd) Not known without mB.

Bμs=0.5

ω=10 s-1

Solution

Energy in SHM

MLook again at the block & spring

energy) mechanical (totalconstant a

)(cos)(sin

)(cos)(sin

!

)(cos

)(sin

221

22221

222221

22212

21

222212

21

kA

ttkA

tktmAUK

k

tkAkxU

tAmmvK

We could also write E = K+U = ½ m(vmax )2

U, K oscillate back and forth “out of phase” with each other; the total E is constant.

n.b.! U, K go through two oscillations while the position x(t) goes through one.

U K

E

t

T

x

v

t

1) What happens to the maximum speed?a) Doublesb) 4 x Largerc) Doesn’t change

2) What happens to the maximum acceleration?a) Doublesb) 4 x Largerc) Doesn’t change

3) What happens to the the total energy?a) Doublesb) 4 x Largerc) Doesn’t change

Suppose you double the amplitude of the motion:

Summary

SHM: )cos( tAx

Definitions: amplitude, period, frequency, angular frequency, phase, phase constant.

(get v, a with calculus)

The acceleration is proportional to displacement: a(t) = -2 x(t)

Practice problems, Chapter 15:3, 5, 11, 19, 23

(6th ed – Chapter 13)1, 3, 5, 9, 19, 23