Group 2 T8

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University of British Columbia *

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131

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Electrical Engineering

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Apr 3, 2024

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Tutorial Section: ____TD2____ Breakout Room Number: ____2____ Names Harmony Chao Kate Zaglinska Bryan Wong Maya Geerts Renz Tingson Tutorial 8 Questions Feel free to press the Return key within the answer boxes to take as much (or little) space as you like! Question 1: Speed trap Police RADAR speed detectors bounce microwave light radiation off of moving vehicles and detect the reflected waves. These waves are shifted in frequency by the Doppler effect , and the frequency shift between the ∆𝑓 = 𝑓' − 𝑓 outgoing wave and the received waves provides a measure of the vehicle speed. Though the waves here are light instead of sound, the Doppler shift works exactly the same way, except that becomes 𝑣 = 343 𝑚/? . 𝑐 = 3. 0 × 10 8 𝑚/? a) (1 mark) A stationary police car sends out a microwave of frequency . Write an expression for the frequency 𝑓 𝑓 𝑜 observed by a vehicle moving towards the police car at speed . ( N.B. should be a function of , , and .) 𝑣 𝑓 𝑜 𝑣 𝑐 𝑓 f o =f s (c+v red )/(c) f o =f(c+v)/(c) b) (1 mark) The radar waves are now reflected off of the moving car, so the car moving at speed becomes an 𝑣 emitter of the radar wave (a moving source of the wave). Write an expression for the frequency emitted by the vehicle, which is the frequency of the reflected wave detected by the stationary police car, in terms of , and . 𝑣, 𝑐 𝑓 f o -is the police car here f red = f 0 from previous question f o = f red (c)/(c-v) f o = f(c+v)/(c-v)

c) (0.5 marks) The Doppler shift is what is measured by the radar gun. Find an expression for the Doppler shift , in terms of , , and . ∆𝑓 = 𝑓' − 𝑓 𝑣 𝑐 𝑓 = [ f red (c)/(c-v red )] - [f s (c+v red )/(c)] ∆𝑓 = f(c+v)/(c-v) - f(c+v)/c) ∆𝑓 = f(c+v)/(c-v) - f = f [ (c+v)/(c-v) - 1 ] ∆𝑓 d) (1 marks) Note that the vehicle speed , or equivalently , and so may be neglected 𝑣 << 𝑐 𝑣/𝑐 << 1 𝑣/𝑐 compared to 1. What does your expression for simplify to, in this limit? (Your answer should depend on , , and ∆𝑓 𝑣 𝑐 , and it should not be zero!) 𝑓 = [ f red (c)/(c-v red )] - f s ∆𝑓 f(c+v)/(c) = f[(c/c)+(v)/(c)] = fs [1+(v)/(c)] … . so = f 𝑣/𝑐 << 1 = f(c+v)/(c-v) - f(c+v)/c) ∆𝑓 = f(c+v)/(c-v) - f = f [ (c+v)/(c-v) - 1 ] ∆𝑓 = f [ {(c+v)-(c-v)}/(c-v) ] = f (2v/(c-v)) = f*2v/(c-v) e) (2 marks) For an emitted radar gun frequency of , a radar Doppler shift of 𝑓 = 30 ??? = 30 × 10 9 ?? is observed for a car driving on a highway with a speed limit. Was the car speeding? Calculate 6400 ?? 100 𝑘𝑚/ℎ? the speed. 100 km/hr = 27.78 m/s 6400Hz = (30*10^9)*2v/(3*10^8 - v) v = 31.99m/s m/s * 3.6 = km/h v = 115.2 km/h > 100 km/h Yes, they were speeding. Question 2: Recognize the wave

The figure shows a standing wave on a 8.2 m long string. The string vibrates up and down with 21 complete vibrational cycles in 5 seconds. a) (1 mark) What is the wavelength of the traveling waves on this string? ƛ =8.2/5=1.64*2 = 3.28m b) (1 mark) Determine the frequency and the period of the traveling waves. f =21/5= 4.2Hz T =1/f=1/(4.2)= (5/21)s c) (0.5 marks) What is the speed of the traveling waves? v = ƛ *f=3.28*4.2= 13.78m/s d) (1 mark) The amplitude of the traveling component waves on the string is A . In terms of A , write the wave function, , of the standing wave on the string in the simplest form. ?(?, ?) y(x,t)=2Asin((1.92)x)cos((26.4)t) k =(2pi)/ ƛ =(2pi)/(3.28)= 1.92 w= (2pi)/T=(2pi)/(5/21)= 26.4 e) (2 marks) If the max amplitude of the standing wave is 0.1m, find the maximum amplitude of oscillation, and the maximum speed of the string, at the point . ? = 0. 55 𝑚 x=0.55 t=0 y(x,t)=0.1sin((1.92)*0.55)cos((26.4)0) A max of oscillations = 0.0870393 m v max = A max of oscillations *w = (0.0870393)*(26.4)= 2.2978 m/s velocity=(dy/dt)=-2Awsin(kx)sin(wt) = -2Awsin(kx)=-0.1(26.4)sin(1.92(0.55)) = -2.2978 m/s final answer = 2.2978 m/s

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