57
Week 10 Chapter 15 & 16 Waves and Oscillations 1

Week 10 Waves Ch15 16 FINAL

Embed Size (px)

DESCRIPTION

Physics Waves

Citation preview

Page 1: Week 10 Waves Ch15 16 FINAL

Week 10 Chapter 15 & 16 Waves and Oscillations

1

Page 2: Week 10 Waves Ch15 16 FINAL

Lecture 1:

Oscillations

- types of waves 16.2

- transverse and longitudinal waves 16.3

- wavelength, frequency and speed of wave 16.4, 16.5

Lecture 2:

Waves

- simple harmonic motion 15.2, 15.3

Lecture 3:

Waves

- wave speed on stretched string 16.6

- energy and power 16.7

Week 10

2

Page 3: Week 10 Waves Ch15 16 FINAL

3

β€’ A wave is an oscillation that moves through space or matter

transferring energy from one place to another

β€’ Generally – material returns to its original position as the wave

passes by –there for there is no mass transport.

Waves

Page 4: Week 10 Waves Ch15 16 FINAL

4

Classical waves transfer energy without transporting matter through the

medium. Waves in a medium do not move the material in the medium

from one place to another place; instead the wave's energy travels

through the medium, leaving the material in place, similar to a cork

rising and falling in one place as the wave moves past the cork

cork bobs up and

down at the same

position

Waves – Transfer of Energy

Page 5: Week 10 Waves Ch15 16 FINAL

16.2 Types of Waves

β€’ Mechanical waves. These waves have two central features: They

are governed by Newton’s laws, and they can exist only within a

material medium – NO medium NO wave: water waves, sound

waves, and seismic waves (earthquake).

β€’ Electromagnetic waves. These waves require no material medium

to exist = PURE ENERGY. All electromagnetic waves travel

through a vacuum at the same exact speed c, speed of light = 3 x108

ms-1. Common examples include visible and ultraviolet light, radio

and television waves, microwaves, x rays, and radar (radio) waves.

β€’ Matter waves. These waves are associated with electrons, protons,

and other fundamental particles, atoms and molecules moving at

very high speeds where they acquire wave-like properties. Basis of

quantum mechanics. Louise de Broglie (1882 – 1987) Ξ» =β„Ž

π‘šπ‘£.

5

Page 6: Week 10 Waves Ch15 16 FINAL

In a transverse wave, the

displacement of every such oscillating

element along the wave is

perpendicular to the direction of travel

of the wave

16.3 Transverse and Longitudinal Waves

In a longitudinal wave the motion of

the oscillating particles is parallel to

the direction of the wave’s travel

6

Note – energy of wave inversely

proportional to its energy 𝐸 ∝1

πœ†

Page 7: Week 10 Waves Ch15 16 FINAL

7

l

A

Time/distance dis

pla

cem

ent

Wavelength, l, distance between two crests or troughs –

distance to complete one cycle

Amplitude – maximum displacement

Period, T – time between 2 crests/troughs – time to complete

one cycle

Frequency, f = 1/T in Hertz (Hz) – number of oscillations per second

vs

Page 8: Week 10 Waves Ch15 16 FINAL

8

Velocity of a wave, vs = distance/time = l/T = l.f

Waves

q

l

A

Time/distance

dis

pla

cem

ent 1l = 1 cycle = 360 = 2p

y = Asin(q)=Acos(p/2 – q)

y y A

Angular frequency w = 2p.f angular speed, number of cycles per sec, in

radians per sec = rad.s-1

1 π‘Ÿπ‘Žπ‘‘ = 1800

πœ‹ 1 =

πœ‹

1800

Angular wave number, k = 2Ο€/Ξ»- number of cycles per unit distance

the spatial equivalent to frequency = β€œspatial frequency” of the wave

k = phase change of the traveling wave in terms of rad.m-1

vs

k = w/v = proportionality between w and v

Page 9: Week 10 Waves Ch15 16 FINAL

9

16.4 Wavelength and Frequency Wave Motion

1s

angular frequency = number of cycles per sec = 2πœ‹

𝑇 = 2pf rad.s-1

1m

wavenumber = number of cycles per metre = 2πœ‹

πœ† rad.m-1

T

l

Page 10: Week 10 Waves Ch15 16 FINAL

10

Handbook of Recording

Engineering, 1986, p. 2

16.4 Wavelength and Frequency Wave Motion

y = A.sin.f

f

Phase, f, relationship between

2 waves with same frequency

k = number of 2p cycles per

unit distance

Each full cycle f = 360

Note: height is the same whenever x is

an integral multiple of l

Phase angle at distance, x

f = k.x = (2p/l).x

Page 11: Week 10 Waves Ch15 16 FINAL

11

Handbook of Recording

Engineering, 1986, p. 2

16.4 Wavelength and Frequency Wave Motion

y = A.sin.f

f

Dx

y = Asin2Ο€

Ξ». x = Asin k. x

x xt

w = k.v

Page 12: Week 10 Waves Ch15 16 FINAL

16.4 Wavelength and Frequency Wave Motion

12

360 = 2p

f y

y = A.sin.f

l

A

x dis

pla

cem

ent

If the wave is moving at velocity, v, it moves a distance xt =vt

A

w = k.v

y = Asin2Ο€

Ξ». x = Asin k. x

Page 13: Week 10 Waves Ch15 16 FINAL

16.4 Wavelength and Frequency –wave motion

13

y(x,t) = ym sin (kx Β± wt –fo )

General Form

Phase Constant (radians)

- wt wave traveling to right

+ wt wave traveling to left

used to represent the wave as a function of time, t, at a specific distance, x,

or as a function of distance, x, at a specific time, t.

Page 14: Week 10 Waves Ch15 16 FINAL

14

When 2 sine waves of equal amplitude and frequency propagate

through a medium in opposite directions – combine to form a

Standing Wave. Note: No Energy Transfer with Standing Waves

Wave moving to the right is: A sin(kx - Ο‰t) & the left is: A sin(kx + Ο‰t)

Standing wave – addition of the 2 waves

y(x,t) = A sin(kx - Ο‰t) + A sin(kx + Ο‰t)

Using sin A+sin B = 2sin [(A+B)/2] .cos [(A-B)/2]

y ( x, t ) = 2A cos(Ο‰t) * sin kx

Where 2A cos(Ο‰t) determines how the amplitude varies with time

While sin(kx) determines the standing wave’s shape.

Standing Waves

Page 15: Week 10 Waves Ch15 16 FINAL

15

Standing Waves

Fundamental frequency

Β½ sine waveform = Β½ l

2nd Harmonic

1 sine waveform = 1 l

3rd Harmonic

3/2 sine waveform = 3/2 l

4th Harmonic

2 sine waveform = 2 l

l = 2𝐿

𝑛 where n = 1, 2, 3, .......

n = number of nodes

Page 16: Week 10 Waves Ch15 16 FINAL

16

Standing Waves Examples

atomic

orbitals

standing wave

solution for

electrons in an

atom

Page 17: Week 10 Waves Ch15 16 FINAL

16.4 Wavelength and Frequency

The amplitude ym of a wave is the magnitude

of the maximum displacement of the elements

from their equilibrium positions as the

wave passes through them.

The phase of the wave is the argument

(kx –wt) of the sine function. As the wave

sweeps through a string element at a particular

position x, the phase changes linearly with

time t.

The wavelength l of a wave is the distance

parallel to the direction of the wave’s

travel) between repetitions of the shape of the

wave (or wave shape). It is related to the

angular wave number, k, by

:

The period of oscillation T of a wave is the

time for an element to move through one full

oscillation. It is related to the angular

frequency, w, by

The frequency f of a wave is defined as 1/T

and is related to the angular frequency w by

A phase constant f in the wave function:

y =ym sin(kx –wt+ f). The value of f can be

chosen so that the function gives some other

displacement and slope at x 0 when t 0.

17

Page 18: Week 10 Waves Ch15 16 FINAL

16.4 Wavelength and Frequency: The Speed of a Traveling Wave

As the wave in Fig. 16-7 moves,

each point of the moving wave

form, such as point A marked on a

peak, retains its displacement y.

(Points on the string do not retain

their displacement, but points on the

wave form do.) If point A retains its

displacement as it moves, the phase

giving it that displacement must

remain a constant:

18

Page 19: Week 10 Waves Ch15 16 FINAL

Example, Transverse Wave

19

Page 20: Week 10 Waves Ch15 16 FINAL

20

Example, Transverse Wave, Transverse Velocity, and Acceleration

velocity π‘₯, 𝑑 = 𝑣𝑦 = 𝑑𝑦

𝑑𝑑 = βˆ’πœ”π‘¦π‘šcos π‘˜π‘₯ βˆ’ πœ”π‘‘

y π‘₯, 𝑑 = π‘¦π‘šsin π‘˜π‘₯ βˆ’ πœ”π‘‘

acceleration π‘₯, 𝑑 = π‘Žπ‘¦= 𝑑𝑣

𝑑𝑑 = βˆ’πœ”2π‘¦π‘šsin π‘˜π‘₯ βˆ’ πœ”π‘‘

y, vy & ay vs

vy,m = -wym

ay,m = -w2ym

transverse wave speed

Page 21: Week 10 Waves Ch15 16 FINAL

21

y = Asinf v = -wAcosf

π‘ π‘–π‘›πœ™ =𝑦

𝐴 π‘π‘œπ‘ πœ™ = βˆ’

𝑣

πœ”π΄

𝑠𝑖𝑛2πœ™ + π‘π‘œπ‘ 2πœ™ = 1 =𝑦2

𝐴2 +

𝑣2

πœ”2𝐴2

𝑣2=πœ”2 𝐴2 βˆ’ 𝑦2

Relationship between v, w, A and y

Page 22: Week 10 Waves Ch15 16 FINAL

Example, Transverse Wave, Transverse Velocity, and Acceleration

22

Page 23: Week 10 Waves Ch15 16 FINAL

Chapter 15

Oscillations

23

Page 24: Week 10 Waves Ch15 16 FINAL

15.1 Oscillatory motion

Motion which is periodic in time, that is, motion that

repeats itself in time.

Examples:

β€’ Power line oscillates when the wind blows past it

β€’ Earthquake oscillations move buildings

β€’ Weight oscillating on a spring

β€’ Tuning Fork

β€’ Pendulum

β€’ Piston moving in a car engine

β€’ Guitar String

Sometimes the oscillations are so severe, that the

system exhibiting oscillations break apart. 24

Page 25: Week 10 Waves Ch15 16 FINAL

15.2 Simple Harmonic Motion

β€’ A particle repeatedly moves back and forth about the point x=0 where the

time taken for 1 complete oscillation is the period, T.

β€’ In T the particle travels from x=+xm, to –xm, and then back to its original

position xm.

β€’ The velocity vector arrows are scaled to indicate the magnitude of the

speed of the system at different times. At x=Β±xm, the velocity is zero. 25

mg

m

pendulum

Page 26: Week 10 Waves Ch15 16 FINAL

Frequency of oscillation is the number of oscillations that

are completed in each second.

The symbol for frequency is f, and the SI unit is the Hertz

(abbreviated as Hz).

It follows that (sec)

1)(

THzf =

15.2 Simple Harmonic Motion

26

Page 27: Week 10 Waves Ch15 16 FINAL

Any motion that repeats itself is periodic (= harmonic).

If the motion is a sinusoidal function of time, it is called

simple harmonic motion (SHM).

Mathematically SHM can be expressed as:

)cos()( fw = txtx m

Here,

β€’ xm is the amplitude (maximum displacement of the system)

β€’ t is the time

β€’ w is the angular frequency, and

β€’ f is the phase constant or phase angle

15.2 Simple Harmonic Motion

27

Page 28: Week 10 Waves Ch15 16 FINAL

(a) The displacement of two SHM

systems that are different in

amplitudes, but have the same period.

(b) The displacement of two SHM

systems which are different in periods

but have the same amplitude.

(c) The value of the phase constant

term, f, depends on the value of the

displacement and the velocity of the

system at time t = 0. The

displacement of two SHM systems

having the same period and

amplitude, but different phase

constants.

15.2 Simple Harmonic Motion

28

(a)

(c)

(b)

Page 29: Week 10 Waves Ch15 16 FINAL

For an oscillatory motion with period T,

)()( Ttxtx =

The cosine function also repeats itself when the argument

increases by 2p. Therefore,

fT

T

tTt

pp

w

pw

pww

22

2

2)(

==

=

=

Here, w is the angular frequency, and measures the

angle per unit time. Its SI unit is radians/second. To be

consistent, f must also be in radians.

15.2 Simple Harmonic Motion

29

Page 30: Week 10 Waves Ch15 16 FINAL

15.2 Simple Harmonic Motion

The velocity of SHM

)sin()(

cos()(

)(

fww

fw

=

==

txtv

txdt

d

dt

tdxtv

m

m

The maximum value of velocity is: v = wxm.

The phase shift of the velocity is p/2, making

the cosine to a sine function. The acceleration

of SHM is:

)()(

)cos()(

)sin()(

)(

2

2

txta

txta

txdt

d

dt

tdvta

m

m

w

fww

fww

=

=

==

The acceleration amplitude is: a = - w2xm.

In SHM a(t) is proportional to the displacement but opposite in sign. 30

a(t) maximum where

maximum F applied

a(t) maximum

at bottom of

swing.

Page 31: Week 10 Waves Ch15 16 FINAL

15.3 Force Law for SHM

From Newton’s 2nd law:

kxxmmaF === 2w

SHM is the motion executed by a system subject to a force that is

proportional to the displacement of the system but opposite in sign.

Period T of SHM oscillation =

31

F = -kx

Hooke’s law

k is the spring constant

m

k=w

22 )2( fmmk pw ==

k

mT p2=

m

kf

p2

1=

Page 32: Week 10 Waves Ch15 16 FINAL

15.3 The Force Law for Simple Harmonic Motion

The block-spring system shown on

the right forms a linear SHM

oscillator.

The spring constant of the spring,

k, is related to the angular

frequency, w, of the oscillator:

k

mT

m

kpw 2==

32 mg

m

pendulum

T = 2p𝐿

𝑔

L

Page 33: Week 10 Waves Ch15 16 FINAL

Simple pendulum

β€’ Perpendicular to the string

mg sin q = -ma

a is the acceleration along the arc

sin q β‰ˆ q and x β‰ˆ Lq ,

mg sin q = -ma = mg q

a = -g(x/l) = -(g/l)x

as a = -w2x in SHM then w2 = g/l

β€’ Period of oscillation T

g

L2

2T p=

w

p=

A

F=mg

mg cos q

mg sin q

q

x

T

L

33

Page 34: Week 10 Waves Ch15 16 FINAL

Combination of Springs

34

F

F2 = k2x F1 = k1x

F = k1x + k2x = (k1 + k2) x

F

F1 = k1x1

F2 = k2 x2

==

21 k

1

k

1

xxkF

𝟏

π’Œ =

𝟏

π’ŒπŸ+𝟏

π’ŒπŸ

Page 35: Week 10 Waves Ch15 16 FINAL

35

Combination of Springs

F = T1 + T2 = k1x + k2x = (k1 + k2)x

T1

Equilibrium position

T2

compression

of spring

extension

of spring

Spring

constant, k1

Spring

constant, k2

x

motion of block

Page 36: Week 10 Waves Ch15 16 FINAL

Example: Force law:

36

Page 37: Week 10 Waves Ch15 16 FINAL

Example, force law:

37

Page 38: Week 10 Waves Ch15 16 FINAL

Example, force law:

38

Page 39: Week 10 Waves Ch15 16 FINAL

Example, force law:

39

Page 40: Week 10 Waves Ch15 16 FINAL

Example, force law:

40

Page 41: Week 10 Waves Ch15 16 FINAL

Example, force law:

41

Page 42: Week 10 Waves Ch15 16 FINAL

15.4: Energy in Simple Harmonic Motion (not in exam)

The potential energy of a linear oscillator

is associated entirely with the spring.

fw == tkxkxtU m

222 cos2

1

2

1)(

The kinetic energy of the system is

associated entirely with the speed of

the block.

fwfww === tkxtxmmvtK mm

222222 sin2

1sin

2

1

2

1)(

The total mechanical energy of

the system:

222

m kx2

1mv

2

1kx

2

1KUE ===

42 energy oscillates between KE and PE

Page 43: Week 10 Waves Ch15 16 FINAL

Example, energy in SHM:

Many tall building have mass dampers,

which are anti-sway devices to prevent

them from oscillating in a wind. The

device might be a block oscillating at the

end of a spring and on a lubricated track.

If the building sways, say eastward, the

block also moves eastward but delayed

enough so that when it finally moves, the

building is then moving back westward.

Thus, the motion of the oscillator is out of

step with the motion of the building.

Suppose that the block has mass m = 2.72

x 105 kg and is designed to oscillate at

frequency f = 10.0 Hz and with amplitude

xm = 20.0 cm.

(a) What is the total mechanical energy E

of the spring-block system?

43

Page 44: Week 10 Waves Ch15 16 FINAL

Example, energy, continued:

44

Page 45: Week 10 Waves Ch15 16 FINAL

16.6: Wave Speed on a Stretched String

The speed of a wave, v , along a stretched ideal string

depends only on the tension, t , and linear density of the

string, m , and not on the frequency, f , of the wave.

45

m Wave

driver t = mg N

πœ‡ =π‘šπ‘ π‘‘π‘Ÿπ‘–π‘›π‘”

π‘™π‘ π‘‘π‘Ÿπ‘–π‘›π‘” 𝑣 = 𝜏

πœ‡

Dl

vspeed

Page 46: Week 10 Waves Ch15 16 FINAL

16.6: Wave Speed on a Stretched String

A small string element of length Dl

within the pulse is an arc of a circle

of radius R and subtending an angle

2q at the center of that circle.

A force with a magnitude equal to

the tension in the string, t, pulls

tangentially on this element at each

end

46

t cosq t cosq

t sinq t sinq

tangential q

π‘ π‘–π‘›πœƒ =Δ𝑙2

𝑅 = q

Page 47: Week 10 Waves Ch15 16 FINAL

16.6: Wave Speed on a Stretched String The horizontal components of these forces

cancel, but the vertical components add to form

a radial restoring force . For small angles,

If m is the linear mass density of the string, and

Dm the mass of the small element,

The element has an acceleration:

Therefore, as F = ma

47

t cosq t cosq

t sinq t sinq

tangential q

π‘ π‘–π‘›πœƒ =Δ𝑙2

𝑅 = q

𝑣 =πœ”

π‘˜= 𝜏

πœ‡ πœ‡ = 𝜏

𝑣2

𝑓 =1

πœ†πœπœ‡ πœπ‘˜2=πœ‡πœ”2

Page 48: Week 10 Waves Ch15 16 FINAL

16.6: Wave Speed on a Stretched String

48

R R

Velocity, t2 = v2

Velocity, t1 = v1 Distance

travelled

t2 >t1

Change in

velocity

q

q Dv = v2 –v1

π‘‘π‘–π‘ π‘Žπ‘›π‘π‘’ π‘‘π‘Ÿπ‘Žπ‘£π‘’π‘™π‘™π‘’π‘‘

π‘Ÿ= π‘β„Žπ‘Žπ‘›π‘”π‘’ 𝑖𝑛 π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦

π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦

𝑣.Δ𝑑

π‘Ÿ =

Δ𝑣

𝑣

Δ𝑣

Δ𝑑 = a =

𝑣2

π‘Ÿ

Since v1 and v2 are perpendicular to R, the angles of both

triangles are equal. Consequently the two triangles are

similar, the ratios of corresponding sides are equal.

Page 49: Week 10 Waves Ch15 16 FINAL

49

16.6: Wave Speed on a Stretched String

Vibration frequency of a stretched string depends on:

β€’ The length of the string that is free to oscillate - longer

string lower frequency

β€’ The tension in the string - higher tension higher frequency

β€’ The mass of the string - higher mass lower frequency

𝑓 =1

πœ† 𝜏

πœ‡ =

𝑛

2𝑙

𝜏 . 𝐿

π‘š 𝑓 =

𝑛

2

𝜏

π‘šπΏ l =

2𝐿

𝑛

Page 50: Week 10 Waves Ch15 16 FINAL

16.6: Energy and Power of a Wave Traveling along a String

50

Kinetic energy of string element

at each position depends on the

transverse velocity of the element

Potential energy depends on the

amount by which the string element

is stretched as the wave passes

through it

Snapshot of a travelling wave on a

string at t = 0

Page 51: Week 10 Waves Ch15 16 FINAL

l

T

Energy, El

51

β€’ In one period T, energy has moved one wavelength, l

β€’ Transported Power, P = πΈπœ†

𝑇

𝑷 =𝟏

𝟐 ππ•πŸπ‘¨πŸ

𝝀

𝑻=𝟏

𝟐 ππ•πŸπ‘¨πŸπ’—

16.6: Energy and Power of a Wave Traveling along a String

𝑬𝝀 = 𝟏

𝟐 ππ•πŸπ‘¨πŸ 𝝀

Page 52: Week 10 Waves Ch15 16 FINAL

52

16.6: Energy and Power of a Wave Traveling along a String

𝑷𝝀 =𝟏

𝟐 ππ’˜πŸπ‘¨πŸπ’—π’”π’‘π’†π’†π’…

Mass per unit length

of the string (kg/m)

Angular frequency of

wave (radians/sec)

wave velocity

speed (m/s)

Amplitude of wave (m)

Power (Watts)

Page 53: Week 10 Waves Ch15 16 FINAL

53

16.7: Energy and Power of a Wave Traveling along a String

l

vs

Wave motion

A y

vy

𝐾𝑖𝑛𝑒𝑑𝑖𝑐 πΈπ‘›π‘’π‘Ÿπ‘”π‘¦ = βˆ†πΎπΈ = 1

2π‘šπ‘£π‘¦

2 = 1

2πœ‡ Ξ”π‘₯ 𝑣𝑦

2

π‘ƒπ‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ πΈπ‘›π‘’π‘Ÿπ‘”π‘¦ = βˆ†π‘ˆ =1

2𝐹. 𝑦 =

1

2(π‘š. π‘Ž). 𝑦 =

1

2π‘šπœ”2𝑦2

π‘ƒπ‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ πΈπ‘›π‘’π‘Ÿπ‘”π‘¦ = βˆ†π‘ˆ = 1

2π‘šπœ”2𝑦2 =

1

2 m Dx πœ”2𝑦2

πœ‡ =π‘š

Ξ”π‘₯

Page 54: Week 10 Waves Ch15 16 FINAL

54

16.6: Energy and Power of a Wave Traveling along a String

y = Asin(kx – wt)

βˆ†π‘ˆ = 1

2π‘šπœ”2𝑦2 =

1

2πœ”2𝑦2mDx =

1

2mπœ”2A2sin2(kx-wt)Dx

Wave equation – travelling wave

Potential energy for one wavelength along the string is found by

integrating over dx at t=0

π‘ˆπœ† =1

2 πœ‡πœ”2𝐴2 𝑠𝑖𝑛2 π‘˜π‘₯ 𝑑π‘₯ =

1

2 πœ‡πœ”2𝐴2

1

2π‘₯ βˆ’

1

4π‘˜π‘ π‘–π‘› 2π‘˜π‘₯

0

πœ†πœ†

0

= 1

4πœ‡πœ”2𝐴2πœ†

Total Energy with one wavelength = El = Ul + KEl = 1

2πœ‡πœ”2𝐴2πœ†

Δ𝐾𝐸 =1

2π‘šπ‘£2 =

1

2πœ‡π‘£2Ξ”π‘₯ v = wAcos(kx–wt)

βˆ†πΎπΈ = 1

2mπœ”2A2cos2(kx-wt)Dx πΎπΈπœ† =

1

2 πœ‡πœ”2𝐴2 π‘π‘œπ‘ π‘‘π‘₯

πœ†

0

= 1

4πœ‡πœ”2𝐴2πœ†

Similarly

𝑠𝑖𝑛2 π‘˜π‘₯ = 1

21 βˆ’ π‘π‘œπ‘  π‘˜π‘₯

Page 55: Week 10 Waves Ch15 16 FINAL

16.6: Energy and Power of a Wave Traveling along a String

The average power, which is the average rate at which energy of both kinds

(kinetic energy and elastic potential energy) is transmitted by the wave, is:

55

Velocity of string Kinetic energy of string

ym = A

Page 56: Week 10 Waves Ch15 16 FINAL

Example, Transverse Wave:

56

Page 57: Week 10 Waves Ch15 16 FINAL

57

A string of mass of 20 mg and length of 1000 mm carries a wave

with a period 65 ms and a wavelength 60 cm. What is the average

power transmitted along the string if the waves amplitude is 12

mm?

π‘ƒπœ† =1

2 πœ‡π‘€2𝐴2𝑣

v = l .T -1 w = 2p .T -1

π‘ƒπœ†(π‘Šπ‘Žπ‘‘π‘‘) =1

2 πœ‡4πœ‹2

𝑇2𝐴2πœ†

𝑇

Here A = 12 x10-3 m, l = 0.6 m, T = 6.5 x10-2 s, ΞΌ = 2 x10-3 kg/m

P = 12.42 Watts