Simple Harmonic Motion

(15-1)

In fig.a we show snapshots of a s

Simple Harmon

imple

oscillat

ic

ing

Motio

syst

n (

em

SHM)

.

The motion is periodic i.e. it repeats in time. The time needed to complete

( ) ( ) cosmx t x tω φ= +

one repetition is known as the The number of

repetitions per u

(symbol , units: s ).

nit time is called the f

perio

c

d

rquen

T

( )

1

( ) cos

y (symbol

The displacement of the particle is given by the equation:

Fig.b is a plot of ( ) versus . The quantity is called the of the

motion.

ampli

, unit hertz)

I

e

tud

m

mx

fT

t t x

f

x t x tω φ

=

= +

t gives the maximum possible displacement of the oscillating object

The quantity is called the of the oscillator. It is given by angular fre

the equatio

qu

n:

enc

y

ω2

2 fT

πω π= =

( )( ) cosmx t x tω φ= + The quantity is called the phase angle of the oscillator.

The value of is determined from the displacement (0)

and the velocity (0) at 0. In fig.a ( ) is plotted

versus for 0. ( )

x

v t x t

t x t x

φ

φ

φ

=

= =

( ) ( )

Velocity of SHM

velocity

cos

( )( ) cos sin

The quantity is called the

It expresses the maximum possible value of ( )

amplitude

m

m

m m

m

v

t

dx t dv t x t x t

dt dt

x

v t

ω

ω φ ω ω φ

ω

= = + = − +

It expresses the maximum possible value of ( )

In fig.b the velocity ( ) is plott

v t

v t ed versus for 0.

( ) sinm

t

v t x t

φ

ω ω

=

= −

( ) 2 2

2

( )( ) sin cos

The quantity is called the

Acceleration of SHM:

acceleration ampli .It expresses the maximum

possible value of a( ). In fig.c the accele

t

r

d

at

u e am

m m

m

dv t da t x t x t x

dt dt

x

t

ω ω φ ω ω ω

ω

= = − + = − = −

2

ion a( ) is plotted versus for 0.

( ) cosm

t t

a t x t

φ

ω ω

=

= −

( )

2

2 2

We saw that the acceleration of an object undergoing SHM is:

If we apply Newton's second law we get:

The Force Law for Simple Harmonic Motion

Simple harmonic motion occurs whe

a x

F ma m x m x

ω

ω ω

= −

= = − = −

The force can be wr

n the force acting on

itten as:

where is a constant. If we compare the two expressions for F we have

an object is paroportional

to the disaplacement but opposite in sign. F C

C

x= −

2

:

d anm Cω = →2

mT

Cπ=

Consider the motion of a mass attached to a spring

of srping constant than moves on a frictionless

horizontal floor as shown in the figure.

m

k

The net force on is given by Hooke's law: . If we compare this equation

with the expression we identify the constant C with the sping constant k.

We can then calculate the angular fre

F m F kx

F Cx

= −

= −

quancy and the period .

and 2 2

T

C k m mT

m m C k

ω

ω π π= = = = k

mω = 2

mT

kπ=

Example:

Example:

In the figure we show another type of oscillating system

It consists of a disc of rotational

An Angular

inertia

Simple Harmonic Oscillator; Tors

suspended from a

wire that tw

ion Pendul

ists as t

um

m ro

I

ates by an angle . The wire exerts

on the disc a

This is the angular form of Hooke's law. The constant

is called th

restoring torque

torsion constane of

the wire t

.

τ

θ

κ

κθ= −

τ κθ= −

If we compare the expression for the torque with the force equation

we realize that we identify the constant C with the torsion constant .

We can thus readily determine the angular frequen

F Cx

τ κθ

κ

= −

= −

cy and the period of the

oscillation.

We note that I is the rotational inertia of the disc about an axis that coincides wit

2 2

h

the wire. The angle is given b

C I IT

I I C

T

κω π π

κ

ω

θ

= = = =

y the equation: ( ) ( ) cosmt tθ θ ω φ= +

A simple pendulum consists of a particle of mass suspended

by a string of length from a pivot point. If the mass is

disturbed from its equilibrium position

Th

th

e Simple P

e net forc

endu

e ac

lu

n

m

ti g

m

L

on it is such that the system exectutes simple harmonic motion.

There are two forces acting on : The gravitational force and

the tension from the string. The net torque of these forces is:

g

m

r F Lτ ⊥= − = − sin Here is the angle that the thread

makes with the vertical. If 1 (say less than 5 ) then we can

mg θ θ

θ °≪makes with the vertical. If 1 (say less than 5 ) then we can

make the following approximation: where is

expressed in r

s

adians. Wit

in

h this

θ

θθ θ

°

≈

≪

( )

approximation the torque is:

If we compare the expression for Lmg

τ

τ θ τ≈ −

with the force equation we realize that we identify the constant C

with the term . We can thus readily determine the angular frequency

and the period of the oscillation. ; 2

F

C mgLT

I

Cx

Lmg

TI

ω

ω

= =

−

=

=

2I I

C mgLπ π=

In the small angle approximation we assumed that θ << 1 and used the

approximation: sinθ ≅ θ We are now going to decide what is a “small”

angle i.e. up to what angle θ is the approximation reasonably

accurate?

θθθθ (degrees) θθθθ(radians) sinθθθθ

5 0.087 0.087

10 0.174 0.174

15 0.262 0.259 (1% off)

20 0.349 0.342 (2% off)

Conclusion: If we keep θ < 10 ° we make less that 1 % error

2

2

Physical Pendul

The rotational inertia about the pivot point is equal to

Thus 2 2

A physical pendulum is an extened rigid body that is suspended

from a fixed point O and oscill

um

ates

I mL

I mLT

mgL mgLπ π= =

under the influence of gravity

The net torque sin Here is the distance between

point O and the center of mass C of the suspended body.

If we make the small angle approximation 1, we have:

mgh hτ θ

θ

= −

≪

2L

Tg

π=

If we make the small angle approximation 1, we have:θ ≪

( ) . If we compare the torques with the

force equation we realize that we identify

the constant C with the term . We can thus readily

determine period of the oscillatio

2 2

n.

mgh

F C

I IT

C

x

hmg

T

τ

π

θ

π

≈ −

=

= =

−

2

Here is the rotational

inertia about an axis through O. com

mghI

I I mh= + 2 I

Tmgh

π=

Consider an object moving on a circular path of radius

with a uniform speed . If we project th

Simple

e posit

Harmonic Motion and Unifor

ion of the

moving particle

m Circular

at point

Moti

P on the

on

mx

v

′

( )

x-axis we get point P.

The coordinate of P is: ( ) cos .

While point P executes unifrom circular motion its

projection P moves along the x-axis with simple

harmonic motion.

The speed v of point

mx t x tω φ= +

′

P is equal to . The xω′The speed v of point

( )

P is equal to . The

direction of the velocity vector is along the tangent

to the circular path. If we project the velocity

on the x-axis we get: ( ) sin

The acceleration points al

m

m

x

v

v t x t

a

ω

ω ω φ

′

= − +

�

�

( )2

ong the center O. If we project

along the x-axis we get: a( ) cos

Conclusion: Whether we look at the displacement ,the

velocity, or the acceleration , the projection of uniform

circular m

ma t x tω ω φ= − +�

otion on the x-axis diameter is SHM

The mechanical energy of a SHM

is the sum of its poten

Energy in S

tial and ki

imple Harmonic Motion

netic energies an

d

.

E

U K

( )2 2 21 1Potential energy cos

2 2

1 1 1

mU kx kx t

k

ω φ= = +

( ) ( )

( ) ( )

2 2 2 2 2 2

2 2 2 2

1 1 1Kinetic energy sin sin

2 2 2

1 1Mechanical energy cos sin

2 2

In the figure we plot the potential energy

m m

m m

kK mv m x t m x t

m

E U K kx t t kx

U

ω ω φ ω φ

ω φ ω φ

= = + = +

= + = + + + =

(green line), the kinetic energy

(red line) and the mechanical energy (black line) versus time . While and

vary with time, the energy is a constant. The energy of the oscillating object

t

K

E t U

K E

ransfers back and forth between potential and kinetic energy, while the sum of

the two remains constant

When the amplitude of an oscillating object is reduced due

to the presence of an external force the motion

Damped Simple Harmonic Motion

damp

is said to be

An example is given in the figure. A med.

ass m

attached to a spring of spring constant k oscillates vertically.

The oscillating mass is attached to a vane submerged in a

liquid. The liquid exerts a whose

magnitu

damping force

de is given by t

dF�

he equation: dF bv= −

The negative sign indicates that opposes the motion of the oscillating mass.

The parmeter is called the . The net force on m is:

damping

From Newton's second

consta

la

nt

w we

d

net

F

b

F kx bv= − −

�

2

2

2

2

have:

We substitute with and a with and get

the following differential equation: 0 d x dx

m b kxd

dx d xkx bv ma v

dt d

t t

t

d

−

+ +

=

=

−

2

2

Newton's second law for the damped harmonic oscillator:

0 The solution has the form:d x dx

m b kxdt dt

+ + =

In the picture above we plot ( ) versus . We can regard the above solutionx t t

( )/ 2 ( ) co sbt m

mx t x e tω φ− ′= +

/ 2

In the picture above we plot ( ) versus . We can regard the above solution

as a cosine function with a time-dependent amplitude . The angular

frequency of the damped harmonic oscillator

bt m

m

x t t

x e

ω

−

′

2

2

2

is given by the equation:

1 For an undamped harmonic oscillator the energy

2

If the oscillator is damped its energy is not constant but decreases with time.

If the

k=

m

damping is s

4mE kx

b

mω′ − =

/ 2

2 /

mall we can replace with By doing so we find that:

1( ) The mechanical energy decreases exponentially with time

2

bt m

m m

bt m

m

x x e

E t kx e

−

−≈

( )/ 2 ( ) co sbt m

mx t x e tω φ− ′= +

Example:

Moving

support

If an oscillating system is disturbed and then allowed

to oscillate freely the corresponding angular frequency

is calle

Forced Oscillations and Resonance

nad the The satural f me systrequ em ce ancy l. n a

ω

so

be driven as shown in the figure by a moving support that

oscillates at an arbitrary angular frequency . Such a

forced oscillator oscillates at the angular frequency

of the driving force. The

d

d

ω

ω

displacement is given by: of the driving force. The

( )

displacement is given by:

The oscillation amplitude varies

with the driving frequancy as shown in the lower figure.

The amplitude is approximatetly greatest when

( ) co

This condition

s m

d

mx t x xt

ω

φ

ω

ω

=

′= +

is called All mechanical structures

have one or more natural frequencies and if a structure is

subjected to a strong external driving force whose frequency

matches one of the natura

resona

l fre

nce.

quencies, the resulting oscillations

may damage the structure