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