511_PartUniversity Physics Solution

# 511_PartUniversity - Sound and Hearing 16.75 16-19(a IDENTIFY and SET UP Use Eq(15.1 to calculate v 1482 m/s EXECUTE = = = 0.0674 m f 22.0 103 Hz(b

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Sound and Hearing 16-19 16.75. (a) IDENTIFY and SET UP: Use Eq.(15.1) to calculate . λ EXECUTE: 3 1482 m/s 0.0674 m 22.0 10 Hz v f == = × (b) IDENTIFY: Apply the Doppler effect equation, Eq.(16.29). The Problem-Solving Strategy in the text (Section 16.8) describes how to do this problem. The frequency of the directly radiated waves is S 22,000 Hz. f = The moving whale first plays the role of a moving listener, receiving waves with frequency L . f The whale then acts as a moving source, emitting waves with the same frequency, SL f f = with which they are received. Let the speed of the whale be W . v SET UP: whale receives waves (Figure 16.75a) EXECUTE: LW vv = + LS S S ff f v ⎛⎞ ++ ⎜⎟ + ⎝⎠ Figure 16.75a SET UP: whale re-emits the waves (Figure 16.75b) EXECUTE: SW = − L S v f + +− Figure 16.75b But f f ′′ = so WW S . v f v v v −− Then W S W SLS S W 2 1. vv vv f v fff f f Δ= − = = = 4 2(2.20 10 Hz)(4.95 m/s) 147 Hz. 1482 m/s 4.95 m/s f −× Δ= = EVALUATE: Listener and source are moving toward each other so frequency is raised. 16.76. IDENTIFY: Apply the Doppler effect formula L S . f f + = + In the SHM the source moves toward and away from the listener, with maximum speed p p . A ω SET UP: The direction from listener to source is positive. EXECUTE: (a) The maximum velocity of the siren is PP 2. Af A ωω = You hear a sound with frequency () Ls i r e n S , f fv =+ where S v varies between 2 f A π + and f A Lm a x s i r e n P P 2 f v f A =− and i n s i r e n f v f A (b) The maximum (minimum) frequency is heard when the platform is passing through equilibrium and moving up (down). EVALUATE: When the platform is moving upward the frequency you hear is greater than siren f and when it is moving downward the frequency you hear is less than siren . f When the platform is at its maximum displacement from equilibrium its speed is zero and the frequency you hear is siren . f 16.77. IDENTIFY: Follow the method of Example 16.19 and apply the Doppler shift formula twice, once with the insect as the listener and again with the insect as the source. SET UP: Let bat v be the speed of the bat, insect v be the speed of the insect, and i f be the frequency with which the sound waves both strike and are reflected from the insect. The positive direction in each application of the Doppler shift formula is from the listener to the source.

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16-20 Chapter 16 EXECUTE: The frequencies at which the bat sends and receives the signals are related by bat insect bat Li S insect bat insect . vv ff f ⎛⎞ ++ + == ⎜⎟ −− ⎝⎠ Solving for insect , v () Sb a t L bat L bat S bat insect L bat S bat a t Lb a t 1 . 1 fvv v ⎡⎤ + ⎢⎥ + + + + + ⎣⎦ Letting Lr e f l a t and f f gives the result. (b) If bat 80.7 kHz, f = refl 83.5 kHz, f = and bat 3.9 m s, v = then insect 2.0 m s. v = EVALUATE: refl bat f f > because the bat and insect are approaching each other.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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511_PartUniversity - Sound and Hearing 16.75 16-19(a IDENTIFY and SET UP Use Eq(15.1 to calculate v 1482 m/s EXECUTE = = = 0.0674 m f 22.0 103 Hz(b

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