Waves 2006 Physics 23

Armen KocharianLecture 3: Sep 12. 2006

Last TimeWhat is a wave?

A "disturbance" that moves through space.Mechanical waves through a medium.

Transverse vs. Longitudinale.g., string vs. sound

Sinusoidaleach particle in the medium undergoes SHM.

Last Time (cont.)Wave function y(x,t)

displacement (y) as a function of position along the direction of propagation (x) and time (t).

x

y P

Wave moves with velocity v in +ve x-direction

y(x-vt)Do not confuse velocity of wave with velocity of

y(x-vt)

At time t, an element of the string (P) at some x has the same y position as an element located at x-vt at t=0 (Q).y(x,t)=y(x-vt,0)=y(x-vt)

x

y

vP

v

x

y

xPPulse at t=0y(x,t)=y(x,0)

Pulse at later time t

vt

Q

Last time (cont.)Sinusoidal Wave:

Wave Equation:

Energy ConsiderationsWaves carry energy.Think of a pulse on a string.

Energy is transferred from hand to string.Kinetic energy moves down the string.

Consider an elementof a string in motion (left to right):

It moves (accelerates) because each piece of themedium exerts a force on its neighboring piece.What is the force on "a" ?

What is the power?

Power:

This is the rate at which work is being done(P=W/t), and the rate at which energy travelsdown the string.

For sinusoidal wave

Then and masslength

Propertyof the string

Propertyof the wave

This is a general property of mechanical waves:1.Power proportional to square of amplitude2.Power proportional to square of frequency

Maximum value of power:

Average value of power

Since average of sin2 is 1/2

Wave reflection

When a wave encounters an "obstacle", .i.e., a "change in the medium"something happens.For example:

a sound wave hitting a wall is "reflected"a light wave originally traveling in air when it reaches the surface of a lake is partially "reflected" and partially "transmitted".

Wave reflection, stringImagine that one end of the string is held fixed:

Why is the reflected pulse inverted?

When the top of the pulse arrives, the string exerts a downward force on support.Newton's 3rd law support exerts upward force on string.Support-to-string force: downward then upward.

Opposite order as pulse creation (was upward then downward).

Reflected pulse is inverted

Pulse was initially created with upward and then downward force on end of string.When pulse arrives at fixed end, string exerts upward force on support.The end of the string does not move (it is fixed!).By Newton's 3rd law, support exerts downward force on string.

In this (idealized) situation the reflected wave has the same amplitude (magnitude) and velocity (magnitude) as the incoming wave.No energy is lost in the reflection.

Now imagine that one end of the string is free:

Why is the reflected pulse not inverted?Pulse was initially created with upward and then downward force on the far end of the string.When pulse first arrives at free end, there is an upward force on the end of the string.When the top of the pulse arrives, the direction of the force becomes downward.upward and then downward force on the free end of the string

Forces on free end like at far end where the pulse was 1st generated.

No inversion on reflection

Boundary Conditions

The properties ("conditions") at the end of the string (or more generally where the medium changes) are called "boundary conditions".This is jargon, but it is used in many places in physics, so try to remember what it means.

Interferencetim

e

Imagine that the incoming pulse is long.Near the boundary at some point we will have a "meeting" of the incoming pulse and the reflected pulse.The deflection of the string will be the sum of the two pulses. (principle of superposition)

Principle of Superposition

When two (or more) waves overlap, the actual displacement at any point is the sum of the individual displacements.

Total displacement First wave Second wave

Consequence of the fact that wave equation is linear in the derivatives.

Standing Waves

Consider a sinusoidal wave traveling to the left:

String held fixed at x=0 reflected wave:

kx+ωt kx-ωt because reflected wave travels to the right.what about δ?

Must choose it to match the boundary conditions!

Boundary condition: string is fixed at x=0Mathematically y(x=0,t) = 0 at all times t

But y(x=0,t)=0:

X=0 kx=πkx=2π

2A

Nodes

Imagine that string is held at both ends.L=length of the stringNodes at x=0 and x=L

Only standing waves of very definite wavelengths (and frequencies) are allowed

Normal Modes

If you could displace a string in a shape corresponding to one of the normal modes, then the string would vibrate at the frequency of the normal mode

Surrounding air would be displaced at the same frequency producing a pure sinusoidal sound wave of the same frequency.

In practice when you pluck a guitar string you do not excite a single normal mode.

Because you do not displace the string in a perfectly sinusoidal way

The displacement of the string can be represented as a sum over the normal modes (Fourier series).

adding an infinite numbers of termsyou can get the exact shape

How to control the frequency of the normal modes

Longer strings lower frequencies.Cello vs violin

Higher tension (F) higher frequencies.More messive strings lower frequencies.

Normal Modes

15.28:

T=0.25 s T=0.50 s

T=0.75 s

T=1.00 s T=1.25 s

Normal Modes15.29: 4.0 s

6.0 s

15.27: a) The wave form for the given times, respectively, is shown.

6.0 ms5.0 ms4.0 ms

3.0 ms2.0 ms1.0 ms

7.0 ms

1.0 ms 2.0 ms 3.0 ms

4.0 ms 5.0 ms 6.0 ms

7.0 ms

Normal Modes

15.39:

b) Eq. (15.28) gives the general equation for a standing wave on a string:

SW( , ) ( sin ) siny x t A kx tω=ASW = 2A, so A= ASW 2 = (5.60 cm) 2 = 2.80 cm

c) The sketch in part (a) shows that 3( 2)L λ=2 , 2k kπ λ λ π= = Comparison of y(x, t)

given in the problem to Eq. (15.28) gives k = 0.0340 rad cm.So, λ = 2π (0.0340 rad cm) = 184.8 cm L = 3(λ 3) = 277 cm

a)

Normal Modes

d) λ =185 cm, from part (c)50.0 rad s so 2 7.96Hzfω ω π= = =

period T = 1 f = 0.126 sv = f λ = 1470 cm s

e) vy = dy dt = ωASWsin kx cosω tvy,max = ωASW = (50.0 rad s)(5.60cm) = 280 cm s

f)

f3 = 7.96 Hz = 3 f1, so f1 = 2.65 Hz is the fundamental

f8 = 8 f1 = 21.2 Hz; ω8 = 2πf8 =133 rad s

λ = v f = (1470 cm s) (21.2 Hz) = 69.3 cm and k = 2π λ = 0.0906 rad cm.

( , ) (5.60cm) sin ([0.0906 rad cm] ) sin ([133rad s] )y x t x t=

Normal Modes

15.42: The condition that x L=is a node becomes .nk L nπ=

The wave number and wavelength are related by

knλ n = 2π , and so λn = 2L n.

Normal Modes

15.45:

(48.0m s)1 1 2 2(1.50m)2 3.00 m, 16.0Hz. v

LL fλ = = = = =

b) λ3 = λ1 3 = 1.00 m, f2 = 3 f1 = 48.0Hz.

c) λ4 = λ1 4 = 0.75 m, f3 = 4 f1 = 64.0 Hz.

a)

Normal Modes

15.48: a) From comparison with Eq. 15.4( ), A= 0.75 cm, λ = 2

0.400 cm = 5.00 cm,

f = 125 Hz, T = 1f = 0.00800 s and v = λf = 6.25 m s.

c) To stay with a wave front as t increases,

x decreases and so the wave is moving in the x−

d) From Eq. ( )15.13 , the tension is 2F vμ= =

2(0.50 kg m) (6.25 m s) 19.5 N.=

e) Pav = 12 μFω2 A2 = 54.2 W.

- direction.

Normal Modes15.35: The wave equation is a linear equation, as it is linear in the derivatives,

and differentiation is a linear operation. Specifically,∂y∂x

=∂(y1 + y2 )

∂x=

∂y1

∂x+

∂y2

∂x.

Repeating the differentiation to second order in both x and t,

∂2 y∂x2 = ∂

2 y1

∂x2 +∂2 y2

∂x2 , ∂2 y∂t2 = ∂

2 y1

∂t2 +∂2 y2

∂t2 .

1y and 2y are given as being solutions to the wave equation; that is,

∂2 y∂x2 =

∂2 y1

∂x2 +∂2 y2

∂x2 =1v2

⎛ ⎝ ⎜ ⎞

⎠ ∂2 y1

∂t 2 +1v2

⎛ ⎝ ⎜ ⎞

⎠ ∂2 y2

∂t 2

=1v2

⎛ ⎝ ⎜ ⎞

⎠

∂2 y1

∂t 2 +∂2 y2

∂t 2

⎡

⎣ ⎢ ⎤

⎦ ⎥

=1v2

⎛ ⎝ ⎜ ⎞

⎠ ∂2 y∂t 2

Normal Modes15.51: a) The functions

y( x, t ) = A cos ( kx − ω t )vy = dy dt = + Aω sin ( kx − ωt )vy,max = Aω = 2π fA

f =vλ

and v =F

m L( ), so f =1λ

⎛ ⎝ ⎜ ⎞

⎠ FLM

vy,max =2πA

λ⎛ ⎝ ⎜ ⎞

⎠ FLM

b) To double v y, max increase F by factor of 4

Normal Modes15.62: a)

⎪⎪⎩

⎪⎪⎨

⎧

>−<−<+−<−<−−+

−<−

=

LvtxLvtxLvtxLh

vtxLLvtxLhLvtx

txy

)( for 0)(0for )(

0)(for )( )(for 0

),(

P(x, t) = −F∂y(x, t)

∂x∂y(x, t)

∂t=

−F(0)(0) = 0 for (x − vt) < −L

−F(h L)(−hv L) = Fv(h L)2 for − L < (x − vt) < 0

−F(−h L)(hv L) = Fv(h L)2 for 0 < (x − vt) < L−F(0)(0) = 0 for (x − vt) > L

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

d) Thus the instantaneous power is zero except for −L < (x − vt) < L,

where it has the constant value Fv(h L)2.

The wave moves in positive x direction with speed v, so we must replace x with x-vt

(c) From Eq. (15.21):