Ch 20 – Wave Motion and Traveling Waves

Propagation of a Disturbance

All mechanical waves require

· source of disturbance and

· a disturbable medium that can influence itself

There are two type of waves

Transverse Waves: A traveling wave or pulse that causes the elements of the disturbed medium to moved perpendicular to the direction of propagation.

Longitudinal Waves: A traveling wave or pulse that causes the elements of the medium to move parallel to the direction of the propagation.

Longnitudnal & Transverse Waves: OW-B-LT

Point element (the black point) below shows an object in the water where a wave passes by. The object rises up then goes back down as the wave passes.

As you can see below…the distance traveled by a wave from a point is Δx which equals vt.

We can say the distance traveled by a wave in a given about of time is, vt.

So the position of any wave is given by the starting position (x) and the distance traveled after time (vt).

Travels to the right: y(x,t) = f(x – vt)

Travels to the left: y(x,t) = f(x + vt)

Earthquakes are an example of both types of waves.

Primary (P) waves are the longitude waves which travel faster than the transverse or Secondary (S) waves.

The longitudinal waves travel about 7.5 km/s near the surface and the transverse waves travel near the surface through the earth at 4.5 km/s.

Example:

Ocean waves with a crest-to-crest distance of 10.0 m

can be described by the wave function

y(x, t) = (0.800 m) sin[0.628(x – vt)]

where v = 1.20 m/s.

(a) Sketch y(x, t) at t = 0.

(b) Sketch y(x, t) at t = 2.00 s

Note how the entire wave form has shifted 2.40 m in the positive x direction in this time interval.

Sinusoidal Waves

A sinusoidal wave is created by plotting sin θ versus θ.

Crest

(+ max amplitude)

Trough

(- max amplitude)

Wavelength

(length of complete wave, until start of repetition)

frequency = 1/T

measured in hertz

Amplitude = A

Wave # is k

In the form of

y(x, t) = A sin (a x)

If the wave is moving to the right then

y(x, t) = A sin [a (x – vt)]

v = λ / T

k = 2π / λ ; ω = 2π / T

v = ω/k

and the wave function is generally expressed as

y(x, t) = A sin [kx – ωt + φ]

y(x, t) = A sin (kx – ωt)

v = dy/dt

dy/dt = A(-ω) cos(kx – ωt)

v = -ωAcos(kx – ωt)

a = dv/dt

dv/dt = A(-ω)(-ω)(-cos(kx – ωt))

a = -ω²Asin(kx – ωt)

The greatest that sine or cosine can be…is 1, thus

v_max = ωA

a_max = ω²A

Example

A sinusoidal wave on a string is described by

y = (0.51 cm) sin(kx – wt)

where k = 3.10 rad/cm and w = 9.30 rad/s.

How far does a wave crest move in 10.0 s?

Does it move in the (+) or (-) x direction?

y(x, t) = A sin [ k x – ω t + φ]

y(x, t) = 0.51 sin [310 x – 9.30 t]

(3.1 rad/cm = 310 rad/m)

v = ω / k

v = 9.3/310

v = 0.030 m/s

v_ave = ∆x / ∆t

0.030 = ∆x / 10.0

∆x = 0.300 meters (+x-dir)

The Speed of Waves on Strings

v² = F_tension / μ; F = ma = m v²/r

The radial (centripetal) force is

Tension(sinθ)

F_r = F_T sinθ (by one side)

The rope is being pulled on two sides so total radial force is given by

F_r = 2 F_T sinθ

If θ is small (<10°) then sinθ = θ

F_r = 2 F_T θ

2 F_T θ = (m) v²/r

2 F_T θ = (2 μ R θ) v²/r

F_T = ( μ ) v²

v² = F_T / μ

Mass associated with Δs is μΔs

(where μ is mass per unit length)

m = μ (Δs) Δs = 2(R θ)

m = μ (2 R θ)

m = 2 μ R θ

Example

A piano string having a mass per unit length 10.0 ´ 10^–3 kg/m is under a

tension of 1000 N. Find the speed of a wave traveling on this string.

v² = F_T / μ

v² = 1000 / 0.010

v = 316 m/s

Reflection and Transmission

When a rope fixed to an immoveable object reflects off the object, the wave reflects off in the opposite direction, with the negative amplitude.

If the wave continues to another medium, some of the energy will be transmitted to the new different sized medium, and some of the energy will be reflected.

The total energy will remain the same.

(15.3: T = ½kA²)

Newton’s 3^rd Law…for every action there is an opposite and equal reaction.

Why in the case of the high density rope to low density rope is the reflected wave not inverted?

Similar explanation as the rings…the low density (thin) rope resists change much less than the high density rope. Thus the low density rope accelerates up to the new Amplitude, then the downward tension pulls the rope back down.

For the thin rope to the thick rope…the thick rope resists change similarly to the rope fixed to the wall.

How would the above picture be different if instead of being attached to the wall, it was attached to a freely moving ring (free to move up and down with the same amplitude as the wave)?

Ans:

Demo: Waves on a Rope: OW-B-WR

Demo: Wave Machine with Rods: OW-B-WM

Rate of Energy Transfer by Sinusoidal Waves on Strings

ΔK = ½ m v²

ΔK = ½(μΔx)v²

μ is mass per unit length

v is the transverse speed

dK = ½ μ dx [ v² ]

dK = ½ μ dx [ω²A² cos²(kx – ωt)]

dK = ½ μ ω²A² cos²(kx – ωt) dx

@ t = 0

(ωt just shifts the phase, thus if

we’re attempting to find Total

Energy a phase shift is irrelevant)

dK = ½ μ ω²A² cos²(kx) dx

∫dK = ½ μ ω²A² ∫cos²(kx) dx

integrate by parts

K = ½ μ ω²A² [x/2+sin(2kx)/4k]

from 0 to λ

K = ¼ μ ω²A² λ

Using similar derivation from above, we get our Potential energy function as

U = ¼ μ ω²A² λ

So our total energy is

TE = ¼ μ ω²A² λ + ¼ μ ω²A² λ

TE = ½ μ ω²A² λ

P = Work / time

Work = ΔK

TE = ΔK for one full period

P = ΔK / T

P = ½ μ ω²A² λ / T

v = f λ = λ/T

P = ½ μ ω²A² v

Example

A taut rope has a mass of 0.180 kg and a length

of 3.60 m. What power must be supplied to the

rope in order to generate sinusoidal waves having

an amplitude of 0.100 m and a wavelength of

0.500 m and traveling with a speed of 30.0 m/s?

v = f λ

30.0 = f (0.500)

f =60.0 Hz

ω = 2πf

ω = 2π(60)

ω = 120π rad/s

P = ½ μ ω² A² v

P =½(0.18/3.6) (120π)² (0.100)² 30.0

P = 1070 Watts

The Linear Wave Equations

ΣF_y = F_T sinθ_B – F_T sinθ_A

ΣF_y = F_T (∂y/∂x_B) – F_T (∂y/∂x _A) (eq 16.23)

m a_y = F_T (∂y/∂x_B) – F_T (∂y/∂x _A) (eq 16.24)

m (∂²y/∂t²) = F_T (∂y/∂x_B) – F_T (∂y/∂x _A)

μΔx (∂²y/∂t²) = F_T [∂y/∂x _B – ∂y/∂x_A]

(μ/ F_T) ∂²y/∂t² = [∂y/∂x_B – ∂y/∂x_A] / Δx (eq 16.25)

At this point we need to remember that at some point in time, tension at B was or will be the same as the tension at A at the current time, so that Tsin(θ_B + ωt) - Tsinθ_A = 0 at some time, t.

The partial derivative of any function is

Associate f(x + Δx) with ∂y/∂x_B (1^st derivative of point B) and f(x) with ∂y/∂x_A (1^st derivative of point A)

à The derivative of a point is the slope (1^st derivative), and the definition of a partial derivative gives us the 2^nd derivative, thus ∂²y/∂x² ß

(μ/T) ∂²y/∂t² = ∂²y/∂x²

This is the linear wave equation for a string.

y(x, t) = A sin (kx – ωt) (from 16.2)

∂²y/∂t² = (F_T /μ) ∂²y/∂x²

Notes & explanations

T represents the tension which is a force, we should say F_Tension.

ΣF_y = Tsinθ_B – Tsinθ_A (T_B is going up…thus positive)

If θ is small, θ < 10° we know

(always applicable at small amplitudes)

sin θ = tan θ; & tan θ = y / x;

And since our infinitesimal displacements

(sin θ = dy/dx) are changing with time

(& only concerned with components of “y” that are dependant on “x”)

sin θ = ∂y/∂x

∂²y/∂t²

∂²y/∂x²

y(x, t) = A sin (kx – ωt)

∂y/∂t = -ω A cos (kx – ωt)]

∂²y/∂t² = -ω² A sin (kx – ωt)]

y(x, t) = A sin (kx – ωt)

∂y / ∂x = k A cos (kx – ωt)]

∂²y/∂x² = -k² A sin (kx – ωt)]

à à à à à à ∂²y/∂t² =

-ω² A sin(kx – ωt)] =

(sinθ)_max = (cosθ)_max = 1 -ω² A =

ω² / k² =

v² =

(F_T /μ) ∂²y/∂x²

(F_T /μ) -k²A sin(kx – ωt)]

(F_T /μ) -k²A

F_T / μ

F_T / μ

à à à à à à ∂²y/∂t² =

∂²y/∂x² =

∂²y/∂x² =

(F_T /μ) ∂²y/∂x²

(μ/ F_T) ∂²y/∂t²

(1/v²) ∂²y/∂t²

Example

Show that the wave function

y = ln[b(x – vt)] is a solution to

∂²y/∂x² = (1/v²) ∂²y/∂t², where b is a constant.

y = ln[b(x – vt)]

∂y/∂t = -bv / b(x-vt)

∂²y/∂t² = -1(-bv)(bv) / b²(x-vt)²

∂²y/∂t² = v² / (x-vt)²

y = ln[b(x – vt)]

∂y/∂x = -b / b(x-vt)

∂y/∂x = -1 / (x-vt)

∂²y/∂x² = -1 / (x-vt)²

(1/v²) ∂²y/∂t² = ∂²y/∂x²

Yes, it satisfies the equation

Ch 20 continued – Sound Waves

Speed of Sound Waves

Type of sound wave:

à Audible à Infrasonic à Ultrasonic

The speed of sound is given below. The first is the speed of sound in air. The latter is the speed of sound in liquids and solids.

Find v_sound in mercury, with a bulk modulus of about 2.80 ´ 10¹⁰ N/m² and a density of 13600 kg/m³.

Find v_sound in an Aluminum rod, which has a Young’s modulus of about 7.0 ´ 10¹⁰ N/m² and a density of 2700 kg/m³.

Demo: Flaming Standing Waves: OW-D-SF

Periodic Sound Waves

The harmonic position function is

s (x, t) = s_max cos (kx – ωt)

s_max à maximum position from the equilibrium position or displacement amplitude of the wave

Note: some student still believe k is the spring constant, but no…it’s the wave number

Variation in gas pressure, ΔP, is also periodic

To the right…when x = 0, we are at s_max and pressure density is at equilibrium, so DP = 0

ΔP = ΔP_max sin (kx – ωt)

ΔP_max is the pressure amplitude

ΔP_max = ρvωs_max

P = F/A = ma/A = m (Dv/t) / A

P = r V (Dv/t) / A

P = r(A*s_max)(Dv/t) / A

P = r s_max Dv/t

P = r s_max Dv w

An ultrasonic tape measure uses frequencies above 20 MHz to determine dimensions of structures such as buildings. It does this by emitting a pulse of ultrasound into air and then measuring the time for an echo to return from a reflecting surface whose distance away is to be measured. The distance is displayed as a digital read-out. For a tape measure that emits a pulse of ultrasound with a frequency of 22.0 MHz, (a) what is the distance to an object from which the echo pulse returns after 24.0 ms when the air temperature is 26^oC? (b) What should be the duration of the emitted pulse if it is to include 10 cycles of the ultrasonic wave? (c) What is the spatial length of such a pulse?

v_sound = 331(1+T_c/273)^1/2

v_sound = 331(1+26/273)^1/2

v_sound = 346 m/s

(a) v_ave = Δx / Δt; d = 2x (echo)

346 = 2d / 0.024

d = 4.15 meters

v_ave = Δx / Δt

v_ave = 10λ / Δt

v = f λ

v = 22x10⁶ λ

(c)

L = 10 ( λ )

L = 10 (v/f)

L = 10 (346/22x10⁶)

L = 0.157 mm

(b)

10λ / Δt = 22x10⁶ λ

Δt = 4.54 x 10^-7 sec

Intensity of Periodic Sound Waves

The piston transmits energy to the element of air in the tube. This energy is propagated away from the piston by the sound wave.

To determine the total kinetic energy in one wavelength, we must determine the velocity

v = dx/dt but position, s(x,t) = s_max cos(kx – ωt); is dependant upon x & t, so we must take the partial derivative with respect to time.

v(x,t) = ∂ s(x,t) / ∂t

v(x,t) = ∂ s_max cos(kx – ωt) / ∂t

v(x,t) = -ω s_maxsin(kx - ωt)

ΔK = Δ½ mv²

ΔK = ½ Δm (ω s_max)²sin²(kx - ωt)

If t is constant (at a given instant in time)

ΔK = ½ ΔρV(ω s_max)²sin²(kx)

ΔK = ½ΔρAΔx (ω s_max)²sin²(kx)

dK = ½ρA (ω s_max)² ∫sin²(kx) dx from 0 to λ

K _λ = ¼ ρA(ωs_max)²λ

As with gravitation potential energy, the total potential energy for one wavelength is the same as the kinetic and the total mechanical energy is

E _λ = K _λ +U _λ = ½ ρA(ωs_max)²λ

And the rate of energy transfer is the power of the wave which is the energy that passes by a given point during one period of oscillation

P = Work/time

P = E _λ / T

P = ½ ρA(ωs_max)² (λ / T)

P = ½ ρA(ωs_max)² (v)

P = ½ ρAv(ωs_max)²

How would we define intensity?

Suppose you take a large magnifying glass, go outside on a nice sunny day near June 20^th and play with ants. Would agree the intensity of the light rays that strike the ants is intense?

The intensity I of a wave is defined as the power per unit area,

I = Power / A = Power / 4πr²

This is the rate at which the energy being transported by the wave transfers through a spherical shell unit of radius, r, and area, A, perpendicular to the direction of the wave. The intensity decreases in proportion to the square of the distance from the source.

I = Power / A

I = ½ ρAv(ωs_max)² / A

I = ½ ρv(ωs_max)²

In terms of pressure amplitude (ΔP_max = ρvωs_max)

I = ΔP² / 2ρv

A logarithmic scale is best used to determine the intensity level, β = 10 log (I / I_o)

I₀, 1.00 x 10^-12 W/ m², is the reference intensity also the threshold of hearing at 1000 Hz

Threshold of pain	Threshold of hearing
I = 1.00 W/m^2	I0 = 1.00 x 10^-12 W/ m^2
β = 120 dB	β = 0 dB

Example

What is the sound level that corresponds to an intensity of 10 x 10^-7 W/m² ?

β = 10 log ( I / I_o )

β = 10 log (10 x 10^-7 W/m² / 1.0 x 10^-12 W/m²)

β = 10 log 10⁶ Rule of thumb: A doubling in the loudness is

β = 60 dB approximately equivalent to an increase of 3 dB

Demo: Range of Hearing: OW-C-RH

Example

A loudspeaker is placed between two observers who are 30 m apart, along the line connecting them. If one observer records a sound level of 60.0 dB, and the other records a sound level of 66.0 dB, how far is the speaker from each observer?

Derive the equation

β₂ – β₁ = 20 log (r₁ / r₂) from

β = 10 log (I / I₀)

β₂ – β₁ = 20 log (r₁ / r₂)

66 – 60 = 20 log (r₁ / r₂)

0.3 = log (r₁ / r₂)

2 r₂ = r₁

10 and 20 meters

β₁ = 10 log (I₁ / I₀);

β₂ = 10 log (I₂ / I₀)

β₂ - β₁ = 10 log(I₂/I₀) - 10 log(I₁/I₀)

β₂ - β₁ = 10 [log (I₂/I₀) - log(I₁/I₀)]

β₂ - β₁ = 10 log [ (I₂ / I₀) / (I₁ / I₀)]

β₂ - β₁ = 10 log ( I₂ / I₁ )

P₂ = P₁

β₂ - β₁ = 10 log (P₂/4πr₂² / P₁/4πr₁²)

β₂ - β₁ = 10 log ( r₁ / r₂) ²

β₂ - β₁ = 20 log ( r₁ / r₂)

The Doppler Effect

The Doppler effect is the apparent change in frequency (or wavelength) that occurs (because of motion of the source or observer) of a wave.

When the relative speed of the source and observer is higher than the speed of the wave, the frequency appears to increase (if lower, then freq appears to decrease)

We also call an appearance of increased frequency, a blue shift or blue-shifted…and decreased frequency, a red shift. Why?

ß low energy, freq high energy, high freq à

Radio micro IR R OYGB IV UV X γ

A convenient graphical representation is to use circular arcs concentric to the source which the fronts of which are called wave fronts

When the observer moves toward the source, the speed of the waves relative to the observer is

Actual wavelength doesn’t change so

f’ / v’ = f / v

f’ = f (v’/ v)

Toward: v ’ = v + v_o v_o is the observer speed

The frequency heard by the observer appears higher when the observer approaches the source

f’ = f ( v’ ) / v

f’ = f (v + v_o)/ v

Away: v ’ = v - v_o v_o is the observer speed

The frequency heard by the observer appears lower when the observer moves away from the source

f’ = f ( v’ ) / v

f’ = f (v - v_o)/ v

Example

At the Winter Olympics, an athlete rides her luge down the track while a bell just above the wall of the chute rings continuously. When her sled passes the bell, she hears the frequency of the bell fall by the musical interval called a minor third. That is, the frequency she hears drops to five sixths of its original value. (a) Find the speed of sound in air at the ambient temperature, 10.0°C. (b) Find the speed of the athlete.

(a) v = 331 + 0.6*T_C

v = 331 + 0.6(-10); v = 325 m/s

(b) f’’ / f’ = 5/6

Approaching bell

f’ = f (v + v_o)/ v

Leaving bell

f’’ = f (v - v_o)/ v

5 / 6 = f (v - v_o)/ v / f (v + v_o)/ v

5 / 6 = (v - v_o) / (v + v_o)

5(v + v_o) = 6 (v - v_o)

v_o = 29.5 m/s

The speed of the source can exceed the speed of the wave.

The envelope of these wave fronts is a cone whose apex half-angle is given by

sin θ = vt / v_sourcet

θ = sin^-1(v / v_source)

(θ is the Mach angle)

The ratio v_source / v is referred to as the Mach number

The conical wave front produced when v_source > v is known as a supersonic “shock wave”

The shock wave carries a great deal of energy concentrated on the surface of the cone

There are correspondingly great pressure variations

Demo: Doppler Ball - OW-B-DB

Digital Sound Recording

These two last sections are directly from the publishers power point presentations

n Encoding sound waveforms began as variations in depth of a continuous groove cut in tin foil wrapped around a cylinder

n Sound was then recorded on cardboard cylinders coated with wax

n Next were disks made of shellac and clay

n In 1948, plastic phonograph disks were introduced

n Disadvantages of phonograph records

n The recording quality diminishes with each playing as small pieces of the plastic are worn away or broken by the needle

n The natural irregularities in the plastic produce noise

n The noise is particularly noticeable during quiet periods with high frequencies playing

n In digital recording of sound, information is converted to binary code

n The waveforms of the sound are sampled

n During the sampling, the pressure of the wave is sampled and converted into a voltage

n The graph below shows the sampling process

n These voltage measurements are then converted to binary numbers (1’s and 0’s)

n Binary numbers are expressed in base 2

n Generally, the voltages are recorded in 16-bit “words”

n Each bit is a 1 or a 0

n The strings of ones and zeroes are recorded on the surface of the compact disc

n There is a laser playback system that detects lands and pits

n Lands are the untouched regions

n They are highly reflective

n Pits are areas burned into the surface

n They scatter light instead of reflecting it

n The binary numbers from the CD are converted back into voltages

n The waveform is reconstructed

n Advantages

n High fidelity of the sound

n There is no mechanical wear on the disc

n The information is extracted optically

Motion Picture Sound

n Early movies recorded sound on phonograph records

n They were synchronized with the action on the screen

n Then a variable-area optical soundtrack was introduced

n The sound was recorded on an optical track on the edge of the film

n The width of the track varied according to the sound wave

n A photocell detecting light passing through the track converted the varying light intensity to a sound wave

n Problems

n Dirt or fingerprints on the track can cause fluctuations and loss of fidelity

n Cinema Digital Sound (CDS)

n First used in 1990

n No backup

n No longer used

n Introduced the use of 5.1 channels of sound:

n Left, Center, Right, Right Surround, Left Surround and Low Frequency Effects (LFE)

n Dolby Digital

n 5.1 channels stored between sprocket holes on the film

n Has an analog backup

n First used in 1992

n Digital Theater Sound (DTS)

n 5.1 channels stored on a separate CD

n Synchronized to the film by time codes

n Has an analog backup

n First used in 1993

n Sony Dynamic Digital Sound (SDDS)

n Eight full channels

n Optically stored outside the sprocket holes on both sides of the film

n Both sides serve as a redundancy

n Analog optical backup

n The extra channels are a full channel LFE plus left center and right center behind the screen

This problem represents a possible (but not recommended) way to code instantaneous pressures in a sound wave into 16-bit digital words. Example 17.2 showed that the pressure amplitude of a 120-dB sound is 28.7 N/m². Let this pressure variation be represented by the digital code 65,536. Let zero pressure variation be represented on the recording by the digital word 0. Let other intermediate pressures be represented by digital words of intermediate size, in direct proportion to the pressure.

(a) What digital word would represent the maximum pressure in a 40 dB sound?

(b) Explain why this scheme works poorly for soft sounds.

(c) Explain how this coding scheme would clip off half of the waveform of any sound, ignoring the actual shape of the wave and turning it into a string of zeros. By introducing sharp corners into every recorded waveform, this coding scheme would make everything sound like a buzzer or a kazoo.

For a 40-dB sound,

β = 10 log (I / I₀)

40 = 10log(I/10^-12)

I = 10^-8 W/m²

ΔP_max² = 2 ρ v I

ΔP_max² = 2(1.2)343(10^-8)

ΔP_max = 2.87 x 10^-3 N/m²

(a)

Code = (2.87 x 10^-3/ 28.7)*65536 = 6.5536

Which is a digital word of 7

(b) For sounds of 40 dB or softer, too few digital words are available to represent the wave form with good fidelity.

(c)

In a sound wave ΔP is negative half of the time but this coding scheme has no words available for negative pressure variations.