Beats and Doppler Chapter 17 & 18

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# Beats and Doppler Chapter 17 & 18 - PowerPoint PPT Presentation

Beats and Doppler Chapter 17 &amp; 18. PHYS 2326-28. Concepts to Know. Beats Doppler Effect Sign Convention. Beats.

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### Beats and Doppler Chapter 17 & 18

PHYS 2326-28

Concepts to Know
• Beats
• Doppler Effect
• Sign Convention
Beats
• Listening to a band or orchestra of relative beginners, you’ll notice the problem that the music just doesn’t sound good. If you even listen to two people playing the same thing it could sound bad as well.
• Quite often this is due to tuning of the musical instruments. If one is at a different pitch from the other there is a beat that may be fairly low to very high speed.
Beats
• This beating ruins the effect of the music.
• It is caused by the instruments playing at slightly different pitches and the intensity changes with time rather than with space.
• Beating is the periodic variation in amplitude caused by the superposition of two waves of slightly different frequencies
• The frequency of the beat equals the difference in frequency of the two sources
Beats
• Section 18.7 shows that given two wave functions of different frequencies can be superimposed using a trigonometric identity cos a + cos b = 2 cos((a-b)/2)cos((a+b)/2)
• The superposition gives y = 2a cos (2pi ((f1-f2)/2)t) cos(2pi((f1+f2)/2)t) eqn 18.10
• This shows there is an average frequency that is ‘beating’ at the difference frequency.
Beats
• The envelope that beats becomes
• 2Acos(2 pi ((f1-f2)/2)t)
• the beat frequency = (f1-f2) since cosine goes to zero twice every 2pi radians
An Interesting Effect
• Since this qualifies regardless of the actual frequencies and differences, one can achieve a fascinating musical effect under some conditions. Given 2 flutists playing a duet, music can (and has) been written that makes use of these sum and difference of frequencies – so that rather than a irritating beat due to poor tuning, one can achieve what sounds like 3 or 4 instruments playing different notes played at one time with even a different melody line
Doppler Effect
• So far, we’ve dealt with observing waves emitted from and observed from positions at rest
• We know that the waves travel at a speed relative to the medium
• What happens if the transmitter and / or the receiver are traveling relative to the medium?
• We’ve all heard train and car horns shift frequency as they travel past us
Moving Observer

Eqns 17.9 and 17.10 show what happens to the frequency if an observer moves relative to the medium. Since v=f λ

f = v/λ and for a moving observer f’=v’/ λ = (v+vo)/ λ where v is the velocity of the wave and vo is the velocity of the observer

f’ = ((v+vo)/v) f moving towards the source and f’ = ((v-vo)/v) f moving away from the source

Moving Source
• When a source moves, it moves a distance vsT during each period = vs/f and the wavelength is shortened to λ’ = λ-Δλ = λ- vs/f
• The observer hears f’ = v/ λ = v/(v/f - vs/f)
• f’ = (v/(v – vs)) f source moving towards observer and
• f’ = (v/(v + vs)) f for the source moving away
Sound Wave Doppler
• The general equation becomes eqn 17.13
• f’ = ((v+vo)/(v-vs))f

SIGNS – these signs are for the observer or source moving towards the other and for the case where one is moving away from the other the appropriate sign is reversed.

For Electromagnetic Waves

fr = sqrt((c-v)/(c+v)) fs

where c is the speed of light and v the velocity difference between transmitter and receiver. This is the relativistic equation and cannot distinguish between which is in motion wrt the medium or even if there is a medium.

Example 1

A car traveling at 20m/s blows its horn, f=300Hz as it passes a second car at rest. If the speed of sound is 345m/s, find a) the frequency heard in the second car before it passes, b) the frequency heard after it passes, c) the wavelength ahead of the car, d) the wavelength behind the first car, e) If the second car blows its horn after the first passes, what is the beat frequency heard by those in the second car, f) by those in the first car?

Example 1

a) observer – at rest vo=0,f’ = ((v+vo)/(v-vs))f

f’ = ((345+0)/(345-20))300 = 318.5Hz

b) f’=((345+0)/(345-(-20)) = 345/365 = 284.4Hz

c) wavelength = v/f = 345/317.5 = 1.083m

d) wavelength = v/f = 345/284.4 = 1.213m

e) fbeat = fa-fb = 300Hz-284.4Hz = 15.6Hz

f) f’ = ((345+(-20))/(345+0) = 282.6Hz

f-f’ = 300 – 282.6 = 17.4Hz

Example 2

Student skating at 3m/s towards a wall, blows a 500Hz whistle at 100m from the wall and times the echo. a) How far does he travel before hearing the echo, b) what is the echo frequency? c) what is the beat frequency with the whistle? assume v=345m/s

Example 2
• in time t, sound travels vt and skater travels v2t. The distance traveled by the sound is D + D- v2t since the skater is that much closer to the wall when the sound returns. hence 2D = v2t + vt = 2(100) = 345t + 3t = 348t, t= 0.5745 s

x = vt = 3 * 0.5745 = 1.72m

b) f’ = ((v+vo)/(v-vs))f = ((345+0)/(345-3))500= 504.4 Hz heard by the wall and reflected

For the skater f=((345+3)/(345+0))504.4 = 508.8 Hz

c) fbeat = fa-fb = 508.8-500 = 8.8 Hz

Example 3

Hydrogen gas emits spectral lines in the visible in what is called the Balmer series. The first two are the H-alpha = 656.1nm (nice and red) and H-beta 486.1nm (bluish green). a) how fast is a galaxy moving away from us if its beta line is shifted all the way to the alpha wavelength b) what is the fractional change what is the fractional change in wavelength if the star is moving away from us at 0.5c, c) what is the observed shift in the alpha line for a star moving towards the earth at a speed equal to that of the earth around the sun – orbital radius = 1.5E11m

Example 3
• source wavelength = 4.861E-7, c=fsλs

received wavelength = 6.563E-7, c=fr λr

fr = sqrt((c-v)/(c+v))fs, fr/fs = sqrt((c-v)/(c+v))= λs/ λr, square both sides

(λs/ λr)^2 = (c-v)/(c+v), solving for v

v = (1-(λs/ λr)^2)*c/(1+ (λs/ λr)^2) = 0.291c

b) v= 0.5c, fr = sqrt((c-v)/(c+v))fs

fs/fr = sqrt((c+v)/(c-v)) = λr/λs = sqrt(1.5c/0.5c)= 1.732

fractional change in wavelength Δλ/λ= λr/λs-1 = 0.732

c) t = 1 year = 3.156E+7 sec

λs = 6.563E-7 s, 2pi r = vt,

v=2pi r/t = 29863 m/s

Δλ/λs = sqrt ((c-v)/(c+v)) – 1, multiply sqrt by c-v and

get sqrt((c-v)^2/(c^2-v^2)) and v^2 <<c^2 so result

is (c-v)/(c) = 1-v/c – 1=v/c = 29863/3.0E+8 = 9.96E-5

Δλ ~= (v/c) λs

Δλ = 9.96 E-5 * 6.563E-7 = 6.531E-11 m