Chapter 17 Oscillations

17-1 Oscillating Systems

Each day we encounter many kinds of oscillatory motion, such as swinging pendulum of a clock, a person bouncing on a trampoline, a vibrating guitar string, and a mass on a spring.

They have common properties:

1. The particle oscillates back and forth about a equilibrium position. The time necessary for one complete cycle (a complete repetition of the motion) is called the period T.

2. No matter what the direction of the displacement,

the force always acts in a direction to restore the

system to its equilibrium position. Such a force is

called a “restoring force(恢复力 )”.

3. The number of cycles per unit time is called the

“frequency” f.

(17-1)

Unit: period (s)

frequency(Hz, SI unit), 1 Hz = 1 cycle/s

Tf

1

4. The magnitude of the maximum displacement from equilibrium is called the amplitude of the motion.

17-2/3 The simple harmonic oscillator and its motion

1. Simple harmonic motionAn oscillating system which can be described in terms of sine and cosine functionssine and cosine functions is called a “simple harmonic oscillator” and its motion is called “simple harmonic motion”.

2. Equation of motion of the simple harmonic oscillator

Fig 17-5 shows a simple harmonic oscillator, consisting of a spring of force constant K acting on

a body of mass m that slides on a frictionless horizontal surface. The body moves in x direction.

Fig 17-5

x

xm

mF

o

o

Relaxed state

origin is chosen at here

kxFx 2

2

dt

xdax

2

2

dt

xdmkx

02

2

xm

k

dt

xd (17-4)

Eq(17-4) is called the “equation of motion of the simple harmonic oscillator”. It is the basis of many complex oscillator problems.

Rewrite Eq(17-4) as

(17-5)

We write a tentative solution to Eq(17-5) as

(17-6)

xm

k

dt

xd)(

2

2

3. Find the solution of Eq. (17-4)

)cos( txx m

We differentiate Eq(17-6) twice with respect to the Time.

Putting this into Eq(17-5) we obtain

Therefore, if we choose the constant such that

(17-7)

Eq(17-6) is in fact a solution of the equation of motion of a simple harmonic oscillator.

)cos(22

2

txdt

xdm

)cos()cos(2 txm

ktx mm

m

k2

a) :

If we increase the time by in Eq(17-6), then

Therefore is the period of the motion T.

2

2

k

mT

22

(17-8)(17-9)

)cos(])2

([cos txtxx mm

m

k

Tf

2

11

The quantity is called the angular frequency.f 2

b) : is the maximum value of displacement. We call it

the amplitude of the motion.

c) and : The quantity is called phase of the motion.

is called “phase constant （常相位 )”.

and are determined by the initial position and

velocity of the particle. is determined by the system.

mxmx

t t

mx

How to understand ?)cos( txx m

2

0

T

xtmx

o

tx 图

mx

0

x

to

同相

How to compare the phases of two SHOs with same ?

)cos( 111 txx m

)cos( 222 txx m

)()( 12 tt

12

x

to

为其它超前

落后

t

x

o

π 反相

(c) same: , different:

1 2 3 4 5 6t

-1

-0.5

0.5

1

xt

2 4 6 8t

-1

-0.5

0.5

1

xt a

1 2 3 4 5 6t

-1

-0.5

0.5

1

xt

Fig 17-6 shows several simple harmonic motions.

(a) (b)

(c) (a) same: , different:

mx

(b) same: , different:

mx

Fig 17-6

mx

d). Displacement, velocity, and acceleration

Displacement

Velocity

Acceleration

When the displacement is a maximum in either direction, the speed is zero, because the velocity must now change its direction.

)cos( txx m

)sin( txdt

dxv mx

)cos(22

2

txdt

xda mx

)2

cos( txm

)cos(2 txm (17-11)

tx 图

tv 图

ta 图

T

mx

mx

2mx

2mx

x

v

a

t

t

t

mx

mxo

o

o T

T

)cos( txx m

0取π2

T

)2

πcos( txm

)sin( txmv

)πcos(2 txm

)cos(2 txa m

17-4 Energy in simple harmonic motion

1.The potential energy

(17-12)

2.The kinetic energy

1 2 3 4 5 6

0.2

0.4

0.6

0.8

1

)(cos2

1

2

1 222 tkxkxU m

U(t)

K(t)

T/2 T)(sin2

1

)(sin2

1

2

1

22

2222

tkx

txmmvK

m

m

)sin( txmv

0

(17-13)

Fig 17-8(a)

• Fig17-8(a), both potential and kinetic energies oscillate with time t and vary between zero and maximum value of .

• Both U and K vary with twice the frequency of the displacement and velocity.

3. The total mechanical energy E is

2

2

1mkx

2

2

1mkxUKE (17-14)

mxmx

K(x)

U(x)

x

Fig 17-8 (b)

2

2

1)( kxxU

)()( xUExK

E

At the maximum displacement , .

At the equilibrium position , .

Eq(17-14) can be written quite generally as

2

2

1mkxK

2

2

1mkxU 0K

0U

222

2

1

2

1

2

1mx kxkxmvUK

)( 222 xxm

kv mx

)( 22 xxm

kv mx (17-16)

(17-15)

then

or

Sample problem 17-2

In Fig 17-5, m=2.43kg, k=221N/m, the block is stretched in the positive x direction a distance of 11.6 cm from equilibrium and released. Take time t=0 when the block is released, the horizontal surface is frictionless.(a) What is the total energy? (b) What is the maximum speed of the block?(c) What is the maximum acceleration?(d) What is the position, velocity, and acceleration at t=0.215s?

xmo

Fig 17-5

Solution:

(a)

(b)

(c) The maximum acceleration occurs just at the

instant of release, when the force is greatest

(d)

JmmNkxE m 49.1)116.0)(/221(2

1

2

1 22

smkg

J

m

E

m

Kv /11.1

43.2

)49.1(222 maxmax

2maxmax /6.10

43.2

)116.0)(/221(sm

kg

mmN

m

kx

m

Fa m

sradm

k/9536.0

Since at t=0, then

So at t=0.215s

)cos()( txtx m

mxx m 116.0 0)536.9cos(116.0cos)( ttxtx m

msx 0535.0)215.0)(536.9cos(116.0

smtxv mx /981.0sin 222 /87.4)0535.0()/536.9( smmsradxax

Sample problem 17-3

In Fig17-5, m=2.43kg, k=221N/m, when the block

m is pushed from equilibrium to x=0.0624m, and

its velocity , the external force is

removed and the block begins to oscillate on the

horizontal frictionless surface. Write an equation

for x (t) during the oscillation.

smvx /847.0

xm

Fig 17-5v

0

Solution:

:

Setting this equal to , we have

To find the phase constant , we still need to use the information give for t=0:

???,,)()( φω,xφωtcosxtx mm

JJJ

mmNsmkg

kxmvE

302.1430.0872.0

)0624.0)(/221(2

1)/847.0)(43.2(

2

12

1

2

1

22

22

2

2

1mkx m

k

Exm 1085.0

2

mx

sradm

k/54.9

At t=0

tx 图

T

xto

mxx m 0624.0cos)0(

5751.0)0(

cos mx

x 9.54

)9.54()(1 ωtcosxtx m

)9.54360()9.54()( -ωtcosx-ωtcosxtx mm2 So only will give the correct initial velocity.

smvx /847.0

9.54

★(1)

)()( φωtcosxtx m

)sin()( m φωtωxdt

dxtvx Or

sm

sm

smxv mx

/847.0

/847.0

sin)/035.1(sin)0(

9.541 1.3052

forfor

radians33.51.3052

]33.5)/(54.9cos[)109.0()( radtsradmtx

is correct.

★(2)

p

17-5 Applications of simple harmonic motion

1. The torsional oscillator(扭转振子 )Fig 17-9 shows a torsional oscillator.

If the disk is rotated in the horizontal (xy)

plane, the reference line op will move

to the OQ, and the wire oo’ will be

twisted. The twisted wire will exert a

restoring torque on the disk, tending to

return the system to its equilibrium.

o

O’Fixed clamp

Fig 17-9

R

Q

m

m2

For small twist, the restoring torque is

(17-17)

Here is constant ( the Greek letter Kappa ), and

is called torsional constant.

The equation of motion for such a system is

(17-18)

where is the rotational inertia of the disk about z

axis. Using Eq(17-17) we have

I

z

2

2

dt

dII zz

or (17-19)

Eq(17-19) and (17-5) are mathematically identical.

The solution should be a simple harmonic oscillation in the angle coordinate , (17-20) or (17-21)

2

2

dt

dIk

I

k

dt

d

2

2

I

k2

)cos( tm

k

IT 2

A torsional oscillator is also called torsional pendulum(扭摆 ). The Cavendish balance, used to measure the gravitational force constant G, is a torsional pendulum.

The restoring force is:

(17-22)

If the is small,

2. The simple pendulum(单摆 )

Fig(17-10) shows a simple pendulum of length L and particle mass m.

sinmgF

L

xmgmgF

sin LT

m

mg

Fig(17-10)

(17-23)

x

xm

g

L

Lmg

mmT 2

/22 (17-24)

k

k

3. The physical pendulum(复摆 )

Any rigid body mounted so that it can swing in vertical plane about some axis passing through it is called “physical pendulum”.

P

C

d

Mgx

y

Fig(17-11)

In Fig. 17-11 a body of irregular shape is pivoted about a horizontal frictionless axis through P

The restoring torque for an angular displacement is (17-26)For small angular displacement . (17-27) then (17-28)

(a)The rotational inertia can be found from Eq(17-28).

(17-29)

sinMgdz

kmgdz

Mgd

I

k

IT 22

I

2

2

4MgdT

I

sin

Suppose the mass of the physical pendulum were concentrated at one point with distance L from the pivot, it will form a simple pendulum.

(b)

Mgd

I

g

LT 22

Md

IL (17-30)

Center of oscillation ( 振动中心 )”

The resulting simple pendulum would have same period as the original physical pendulum.

P

C

d

Fig(17-11)

O

L

• The center of oscillation is often also called the “center of percussion ”（撞击中心） .

If an impulsive force in the plane of oscillation acts at the center of oscillation, no effect of this force is felt at the pivot point P.

The point O is called the “center of oscillation” of the physical pendulum.

• If we pivot the original physical pendulum from point O, it will have the same period as it does when pivoted from point P.

P

C

d

Fig(17-11)

O

LP

C

O

Sample problem 17-4

A thin uniform rod of mass M=0.112kg and length L=0.096m is suspended by a wire that passes through its center and is perpendicular to its length. The wire is twisted and the rod set oscillating. The period is found to be 2.14s. When a flat body in the shape of an equilateral triangle is suspended similarly through its center of mass, the period is found to be 5.83m.Find the rotational inertia of the triangle about this axis.

Solution:

The rotational inertia about its Cm is

from Eq(17-21)

252

1060.812

mkgml

I rod

or 242 1038.6)( mkgIT

TI rod

rod

triamgletriamgle

2

1

)(triamgle

rod

triamgle

rod

I

I

T

T

Sample problem 17-5

A uniform disk is pivoted at

its rim ( Fig17-12). Find its

period for small oscillations

and the length of the

equivalent simple pendulum.

Solution:

The rotational inertia about

pivot at the rim is

O

C

R

P

R2

3R

2

3

Fig 17-12

From Eq(17-28) with d=R, then

The simple pendulum having the same period has

a length (Eq(17-30))

The center of oscillation of the disk pivoted at P is

therefore at o, a distance below the point of support.

222

2

3

2

1MRMRMRI

g

R

MgR

IT

2

322

RMR

IL

2

3

R2

3

You may check that the period of the pendulum pivoted at O is the same as that pivoted at P.

Torsional Pendulum

(扭摆 )

Torsional pendulum clock

17-6 Simple harmonic motion and uniform circular motion

Simple harmonic motion can be described as a

projection of uniform circular motion along a

diameter of the circle.

1.1. Fig17-14 shows a particle

P in uniform circular motion. 0

QP

'p t

v

r

Fig 17-14

x

y

At a time t, the vector ( ) makes an angle with x axis, and the x component of is

(17-31)

This is of course identical to Eq(17-6) for the displacement of the simple harmonic oscillator. • The radius r corresponding to ;• Point , which is the projection of P on the x axis, executes simple harmonic motion along the x axis.

r

r

op t

)cos()( trtx

mx'p

2.2. The magnitude of the tangent velocity of the point P is , the x component of is

(17-32)

3.3. The centripetal acceleration is , and its x component is (17-33)

Eqs(17-32) and (17-33) are identical with Eqs(17-11) for simple harmonic motion, again with replace by r.

v

)sin()( trtvx

)cos()( 2 trtax

vr

r2

mx

See动画库 \波动与光学夹 \1-07辅助圆

4.4. In Fig17-14， the y projection OQ of at time t, is

(17-34)

So the projection of uniform circular motion along the y direction also gives simple harmonic motion.

On the contrary, the combination of two simple harmonic motions at right angles, with identical amplitude and

frequencies, can form a uniform circular motion.

)2

3cos( tr

r)sin()( trty

See动画库 \波动与光学夹 \1-10垂直振动的合成

17-7 Damped (阻尼 ) harmonic motion

Up to this point we have assumed that no frictional

force act on the system.

For real oscillator, there may be friction, air resistance act on the system, the amplitude will decrease.

1.1. This loss in amplitude is called “damping” and

the motion is called “damped harmonic motion”.

xm v

0 f

1 2 3 4 5 6

-1

-0.5

0.5

1

1 2 3 4 5 6

-1.5

-1

-0.5

0.5

1

1.5

x

x

t

t

(a)

(b)

t

e

Fig 17-16

Fig17-16 compare the motion of undamped and damped oscillators.

(a) When we add a small damping force, the amplitude gradually decreases to zero but the frequency changes by a negligible amount. In this case Eq(17-6) becomes

(17-36)

where is called the “damping time constant” or the “mean lifetime” of the oscillator.

is the time necessary for the amplitude to drop to 1/e of its initial value.

)cos()( / textx tm

(b) When the damping force is not large, the mechanism energy is

(17-37)

Eq(17-37) shows that the mechanical energy of the oscillator decreases exponentially with time.

/22

2

1)( t

m ekxtE

The energy decreases twice as rapidly as the amplitude.

2*. Mathematical analysis

Assume the damping force is , where b is a

positive constant called the “damping constant”.

With , Newton’s second law gives

or

(17-38)

xx mabvkx

02

2

kxdt

dxb

dt

xdm

xx bvkxF

xbv

The solution is (17-39) where (17-40)

(a) If b is negligible, . It is ideal simple harmonic oscillation.

)'cos()( 2

textx m

bt

m

2' )2

(m

b

m

k

./2 bm

'

If , that is , damping slows downthe motion. This case is called underdamping (欠阻尼 )

Comparing Eqs(17-39) and (17-36) we have

kmb 2 '

(b) When , , the motion decays exponentially to zero with no oscillation at all.

This condition is called “critical damping (临界阻尼 )”. The lifetime has its smallest possible value, .

kmb 2

0'

/1

ot

x 三种阻尼的比较

)(b

(a)

(c)

(c) When , the motion also decays exponentially to zero with no oscillation, called overdamping （过阻尼） .

kmb 2

Forced oscillations:

Oscillations of a system carried out under the action of an external periodical force, such as

or a successive action of an external non-periodical force.

17-8 Forced oscillations and resonance

tFtF mx''sin)(

and , ,Which frequency will the forced oscillation system take?

''

Forced oscillation system takes the frequencyof the external force, namely .''

Resonance: The amplitude of the forced oscillation can increase much as approaches .

'' This condition is known as “resonance” and is called “resonant angular frequency”.

''

Am

plit

ud

e

Fig 17-19

Small damping

Large damping

5.0 25.1''

In the case with damping, the rate at which energy is provided by the driving force exactly matches the rate at which energy is dissipated by the damping force.

String theory: Theory of everything (万物之理 )

Theory of anything (任意之理 )

Everything or nothing?Where string theory stands today

弦理的根本问题是它很难或不可能得到实验证明

Peter Woit (Princeton)

弦论根本不是理论Not even wrong!

简直比错还差！