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Lo5 - The Doppler Effect and Bats

LO5 for Phys 101

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Lo5 - The Doppler Effect and Bats

  1. 1. LO5
  2. 2. What is the Doppler Effect?  Definition: The increase or decrease of frequency of (sound, light, etc) waves due to the relative motion between the source of the wave and its receiver.  For sound waves in particular, represented by the equation:
  3. 3. Components of the Doppler Effect in Sound Waves  Source, s – point that emits the wave at a given...  Speed, vs (m/s) [of the point itself]  Frequency, fs (Hz) [of the wave]  Receiver, r – point at which the emitted wave is received at a given...  Speed, vr (m/s) [of the point itself]  Frequency, fr (Hz) [of the wave]  Speed of sound in air, v = 343m/s
  4. 4. Direction of Motion and the Difference Between +/- and -/+ The signs used in the Doppler equation to represent the speed of the source and receiver depend on their direction of motion relative to one another  +/- : When either the source or receiver or both are moving TOWARDS each other, there is an increase in frequency, and so the equation favors the numerator (larger value; smaller value for the denominator), meaning that “+” is used for vr and “–” is used for vs  -/+ : When either the source or receiver or both are moving AWAY FROM each other, there is a decrease in frequency, and so the equation favors the denominator (larger value; smaller value for the numerator), meaning that “-” is used for vr and “+” is used for vs
  5. 5. What does the Doppler Effect have to do with Bats?  Used as a means to hunt prey  As bats fly, they emit whistles and listen for the echoes as the sound waves reflect off its prey  Brain detects the change in pitch (frequency) between the whistle and the echo to judge the speed of its prey and adjust its own speed accordingly  In this case:  Whistle = source sound wave  Bat = source speed  Reflected sound waves (echo) = received sound wave  Prey = received speed Source: http://www.scienceclarified.com/everyday/Real-Life-Physics-Vol-2/Doppler-Effect-Real-life-applications.html
  6. 6. Important Concepts to Remember on the Doppler Effect  Motion between source and receiver (whether it is one moving or both):  Towards = higher received frequency = +/-  Apart = lower received frequency = -/+  Speed of sound, v = 343m/s
  7. 7. Practice Problem! Barry the bat is on the hunt for some prey. He emits a whistle at a wavespeed of 20m/s and wavelength 2.5m. He is able to detect an echo pitch from his prey as small as 6Hz. a) What is the frequency of Barry’s whistle? b) What is the maximum speed of the prey moving away is Barry able to detect the smallest pitch if he is at rest? c) After emitting a whistle from rest, Barry detects an echo from a nearby unsuspecting prey also at rest. What frequency is the echo received? Explain your results. d) Heading towards his unsuspecting prey at 15m/s, Barry emits a whistle once more. What frequency does he receive from the echo? e) The prey realizes Barry is on the hunt, and quickens its speed to 20m/s to run away. i) Predict whether the echo frequency becomes higher or lower than Barry’s whistle. Explain. ii) Solve for the change in frequency.
  8. 8. Barry the bat is on the hunt for some prey. He emits a whistle at a wavespeed of 20m/s and wavelength 2.5m. He is able to detect an echo pitch from his prey as small as 6Hz. a)What is the frequency of Barry’s whistle? ANSWER: λ = vwave / f So, F = vwave / λ = 20m/s / 2.5m = 8s-1 = 8 Hz
  9. 9. b) What is the maximum speed of the prey moving away is Barry able to detect the smallest pitch if he is at rest? ANSWER: because moving AWAY, -/+ vs = 0m/s fs= 8Hz fr= 6Hz fr = fs (v – vr)/(v + 0) 6Hz = 8Hz (343m/s – vr)/ (343m/s) 2058m-1/8Hz = 343m/s – vr -85.75 m/s = - vr vr = 86 m/s
  10. 10. c) After emitting a whistle from rest, Barry detects an echo from a nearby unsuspecting prey also at rest. What frequency is the echo received? Explain your results. ANSWER: fs= 8 Hz vs= 0m/s vr = 0m/s Therefore, = fs(v +/- 0)/(v+/- 0) = fs (v/v) fr = fs = 8 Hz The frequency received is the same as that emitted, because neither Barry or his prey are moving relative to one another, so the frequency of the reflected sound waves are not influenced by the speed of the wave source/receiver.
  11. 11. d) Heading towards his unsuspecting prey at 15m/s, Barry emits a whistle once more. What frequency does he receive from the echo? ANSWER: moving towards , +/- vr= 0m/s (still unsuspecting prey) vs = 15m/s fs = 8Hz fr = fs (v+0)/(v-vs) = fs v/(v-vs) = 8Hz (343m/s)/(343m/s – 15m/s) fr = 8.4 Hz
  12. 12. e) The prey realizes Barry is on the hunt, and quickens its speed to 20m/s to run away. i) Predict whether the echo frequency becomes higher or lower than Barry’s whistle. Explain. ii) Solve for the change in frequency. ANSWER: i) From part d) we know that Barry is moving at 15m/s. Because the prey is running away at a greater speed than Barry we can assume that the distance between them is becoming greater, indicating a movement AWAY from one another, and thus a lower echo frequency. ii) Because they are moving away from one another, -/+ will be used. fr =fs (v-vr)/(v+vs) = 8Hz (343m/s-20m/s)/(343m/s+15m/s) fr = 7.2 Hz **Asking for the CHANGE in frequency, so = fs-fr = 8Hz – 7.2Hz =0.8Hz
  13. 13. Tips on Approaching the Doppler Effect Problems 1) Identify what you are solving for, and whether such quantity belongs to the receiver or the source. 2) (IMPORTANT) Establish a motion relationship between the source and receiver (+/- if moving towards one another/received frequency is higher than sourced; -/+ if moving away from one another/received frequency is lower than sourced). 3) Substitute known values, including that of v (speed of sound = 343m/s) into the Doppler Equation, and solve for unknowns and their respective units.

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