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Unformatted text preview: This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A stereo speaker represented by P in the fig ure emits sound waves with a power output of 107.0 W. P x b b 14 m What is the intensity of the sound waves at point x , 14.0 m away? Correct answer: 0 . 0434428 W / m 2 . Explanation: Basic Concept: intensity = P 4 πr 2 Given: P = 107 . 0 W r = 14 . 0 m Solution: intensity = 107 W 4 π (14 m) 2 = 0 . 0434428 W / m 2 002 (part 1 of 2) 10.0 points Two loudspeakers are placed on a wall 4 . 92 m apart. A listener stands directly in front of one of the speakers, 79 . 6 m from the wall. The speakers are being driven by the same electric signal generated by a harmonic oscillator of frequency 1070 Hz. The speed of the sound in air is 343 m / s. What is the phase difference ΔΦ between the two waves (generated by each speaker) when they reach the listener? Correct answer: 2 . 97744 rad. Explanation: Let : r 1 = 79 . 6 m , f = 1070 Hz , and v = 343 m / s . Each speaker produces a spherical wave of the form δP ∝ sin( ϕ ) , ϕ = k r − ω t − φ (0) . where r is the straightline distance from the speaker to the listener. The two speakers have equal frequencies ω 1 = ω 2 = ω and equal initial phases φ (0) 1 = φ (0) 2 = φ (0) , but they are at different distances from the listener: The distance to the first speaker is r 1 = 79 . 6 m but the distance to the second speaker is r 2 = radicalBig (79 . 6 m) 2 + (4 . 92 m) 2 = 79 . 7519 m . Consequently, there is a phase difference be tween the two waves Δ ϕ ≡ ϕ 2 − ϕ 1 = parenleftBig k r 2 − ω t − φ (0) parenrightBig − parenleftBig k r 1 − ω t − φ (0) parenrightBig = k ( r 2 − r 1 ) . The wavenumber of each sound wave fol lows from wavelength which in turn follows from the frequency: k = 2 π λ = 2 π f v = 2 π (1070 Hz) 343 m / s = 19 . 6006 rad / m . Consequently, the phase difference is Δ ϕ = k ( r 2 − r 1 ) = (19 . 6006 rad / m) × (79 . 7519 m − 79 . 6 m) = 2 . 97744 rad . homework 40 – Turner – (58185) 1 hinojosa (jlh3938) – homework 40 – Turner – (58185) 2 003 (part 2 of 2) 10.0 points Suppose the waves generated by each speaker have equal pressure amplitudes δP max 1 = δP max 2 = δP max = 0 . 0074 Pa at the listener’s location. What is the pressure amplitude δP max 1+2 of the combined sound of the two speakers? Correct answer: 0 . 00121339 Pa. Explanation: Let : δP max = 0 . 0074 Pa . The two individual waves have pressures δP 1 = δP max sin( ϕ 1 ) , δP 2 = δP max sin( ϕ 2 ) . Adding them up to obtain the combined sound wave, we find δP 1+2 = δP 1 + δP 2 = δP max bracketleftBig sin( ϕ 1 ) + sin( ϕ 2 ) bracketrightBig = δP max 2 cos parenleftbigg ϕ 2 − ϕ 1 2 parenrightbigg × sin parenleftbigg ϕ 2 + ϕ 1 2 parenrightbigg = δP max 2 cos parenleftbigg Δ ϕ 2 parenrightbigg × sin k r 1 + r 2 2 − ω t −...
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 Fall '08
 Turner
 Work, Frequency, m/s, Hz, Hinojosa

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