# Beats and Doppler Chapter 17 18

0
0
2762 days ago, 831 views
PowerPoint PPT Presentation

### Presentation Transcript

Slide 1

﻿Beats and Doppler Chapter 17 & 18 PHYS 2326-28

Slide 2

Concepts to Know Beats Doppler Effect Sign Convention

Slide 3

Beats Listening to a band or symphony of relative apprentices, you'll see the issue that the music simply doesn't sound great. In the event that you even listen to two individuals playing a similar thing it could sound terrible also. Regularly this is because of tuning of the musical instruments. On the off chance that one is at an alternate pitch from the other there is a beat that might be genuinely low to rapid.

Slide 4

Beats This beating ruins the impact of the music. It is brought about by the instruments playing at marginally extraordinary pitches and the power changes with time as opposed to with space. Beating is the intermittent variety in adequacy brought on by the superposition of two rushes of marginally extraordinary frequencies The recurrence of the beat rises to the distinction in recurrence of the two sources

Slide 5

Beats Section 18.7 demonstrates that given two wave elements of various frequencies can be superimposed utilizing a trigonometric character 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 a normal recurrence that is "beating" at the distinction recurrence.

Slide 6

Beats The envelope that beats gets to be 2Acos(2 pi ((f1-f2)/2)t) the beat recurrence = (f1-f2) since cosine goes to zero twice every 2pi radians

Slide 7

An Interesting Effect Since this qualifies paying little respect to the genuine frequencies and contrasts, one can accomplish an intriguing musical impact under a few conditions. Given 2 flute players playing a two part harmony, music can (and has) been composed that makes utilization of these whole and distinction of frequencies – so that as opposed to a bothering beat because of poor tuning, one can accomplish what sounds like 3 or 4 instruments playing diverse notes played at one time with even an alternate tune line

Slide 8

Doppler Effect So far, we've managed watching waves radiated from and seen from positions very still We realize that the waves go at a speed in respect to the medium What happens if the transmitter and/or the recipient are setting out in respect to the medium? We've all heard prepare and auto horns move recurrence as they go past us

Slide 9

Moving Observer Eqns 17.9 and 17.10 show what happens to the recurrence if an onlooker moves with respect to the medium. Since v=f λ f = v/λ and for a moving spectator f'=v'/λ = (v+v o )/λ where v is the speed of the wave and v o is the speed of the eyewitness f' = ((v+v o )/v) f moving towards the source and f' = ((v-v o )/v) f moving far from the source

Slide 10

Moving Source When a source moves, it moves a separation v s T amid every period = v s/f and the wavelength is abbreviated to λ " = λ - Δλ = λ - v s/f The onlooker hears f' = v/λ = v/( v/f - v s/f) f' = (v/(v – v s )) f source moving towards spectator and f' = (v/(v + v s )) f for the source moving endlessly

Slide 11

Sound Wave Doppler The general condition gets to be eqn 17.13 f' = ((v+v o )/(v-v s ))f SIGNS – these signs are for the eyewitness or source moving towards the other and for the situation where one is moving far from the other the suitable sign is turned around.

Slide 12

For Electromagnetic Waves fr = sqrt((c-v)/(c+v)) fs where c is the speed of light and v the speed distinction amongst transmitter and collector. This is the relativistic condition and can't recognize which is in movement wrt the medium or regardless of the possibility that there is a medium.

Slide 13

Example 1 An auto going at 20m/s blows its horn, f=300Hz as it passes a second auto very still. In the event that the speed of sound is 345m/s, discover a) the recurrence heard in the second auto before it passes, b) the recurrence heard after it passes, c) the wavelength in front of the auto, d) the wavelength behind the primary auto, e) If the second auto blows its horn after the principal passes, what is the beat recurrence heard by those in the second auto, f) by those in the main auto?

Slide 14

Example 1 a) spectator – very still vo=0,f' = ((v+v o )/(v-v s ))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

Slide 15

Example 2 Student skating at 3m/s towards a divider, blows a 500Hz shriek at 100m from the divider and times the resound. a) How far does he go before listening to the resound, b) what is the reverberate recurrence? c) what is the beat recurrence with the shriek? expect v=345m/s

Slide 16

Example 2 in time t, sound ventures vt and skater voyages v 2 t. The separation went by the sound is D + D-v 2 t since the skater is that much nearer to the divider when the sound returns. subsequently 2D = v 2 t + vt = 2(100) = 345t + 3t = 348t, t= 0.5745 s x = vt = 3 * 0.5745 = 1.72m b) f' = ((v+v o )/(v-v s ))f = ((345+0)/(345-3))500= 504.4 Hz heard by the divider 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

Slide 17

Example 3 Hydrogen gas emanates ghostly lines in the unmistakable in what is known as the Balmer arrangement. The initial two are the H-alpha = 656.1nm (pleasant and red) and H-beta 486.1nm (pale blue green). a) how quick is a system moving far from us if its beta line is moved the distance to the alpha wavelength b) what is the partial change what is the fragmentary change in wavelength if the star is moving far from us at 0.5c, c) what is the watched move in the alpha line for a star moving towards the earth at a speed equivalent to that of the earth around the sun – orbital sweep = 1.5E11m

Slide 18

Example 3 source wavelength = 4.861E-7, c=fs λ s got wavelength = 6.563E-7, c=fr λ r fr = sqrt((c-v)/(c+v))fs, fr/fs = sqrt((c-v)/(c+v))= λ s/λ r, square both sides ( λ s/λ r)^2 = (c-v)/(c+v), unraveling for v = (1-( λ s/λ r)^2)*c/(1+ ( λ s/λ r)^2) = 0.291c

Slide 19

b) v= 0.5c, fr = sqrt((c-v)/(c+v))fs fs/fr = sqrt((c+v)/(c-v)) = λ r/λ s = sqrt(1.5c/0.5c)= 1.732 partial change in wavelength Δλ/λ = λ r/λ s-1 = 0.732 c) t = 1 year = 3.156E+7 sec λ s = 6.563E-7 s, 2pi r = vt, v=2pi r/t = 29863 m/s Δλ/λ s = sqrt ((c-v)/(c+v)) – 1, increase 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) λ s Δλ = 9.96 E-5 * 6.563E-7 = 6.531E-11 m