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Unformatted text preview: klare (alk736) – homework 35 – Turner – (58220) 1 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 106.1 W. P x 7 . 18 m What is the intensity of the sound waves at point x , 7.18 m away? Correct answer: 0 . 163778 W / m 2 . Explanation: Let : P = 106 . 1 W and r = 7 . 18 m . I = P 4 π r 2 = 106 . 1 W 4 π (7 . 18 m) 2 = . 163778 W / m 2 . 002 (part 1 of 2) 10.0 points Two loudspeakers are placed on a wall 3 . 52 m apart. A listener stands directly in front of one of the speakers, 32 . 4 m from the wall. The speakers are being driven by the same electric signal generated by a harmonic oscillator of frequency 2820 Hz. The speed of the sound in air is 343 m / s. What is the phase di ff erence ΔΦ between the two waves (generated by each speaker) when they reach the listener? Correct answer: 9 . 84848 rad. Explanation: Let : r 1 = 32 . 4 m , f = 2820 Hz , and v = 343 m / s . Each speaker produces a spherical wave of the form δ P ∝ 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 di ff erent distances from the listener: The distance to the first speaker is r 1 = 32 . 4 m but the distance to the second speaker is r 2 = (32 . 4 m) 2 + (3 . 52 m) 2 = 32 . 5906 m . Consequently, there is a phase di ff erence be tween the two waves Δ ϕ ≡ ϕ 2 ϕ 1 = k r 2 ω t φ (0) k r 1 ω t φ (0) = 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 (2820 Hz) 343 m / s = 51 . 6577 rad / m . Consequently, the phase di ff erence is Δ ϕ = k ( r 2 r 1 ) = (51 . 6577 rad / m) × (32 . 5906 m 32 . 4 m) = 9 . 84848 rad . klare (alk736) – homework 35 – Turner – (58220) 2 003 (part 2 of 2) 10.0 points Suppose the waves generated by each speaker have equal pressure amplitudes δ P max 1 = δ P max 2 = δ P max = 0 . 0049 Pa at the listener’s location. What is the pressure amplitude δ P max 1+2 of the combined sound of the two speakers? Correct answer: 0 . 00206066 Pa. Explanation: Let : δ P max = 0 . 0049 Pa . The two individual waves have pressures δ P 1 = δ P max sin( ϕ 1 ) , δ P 2 = δ P max sin( ϕ 2 ) . Adding them up to obtain the combined sound wave, we find δ P 1+2 = δ P 1 + δ P 2 = δ P max sin( ϕ 1 ) + sin( ϕ 2 ) = δ P max 2 cos ϕ 2 ϕ 1 2 × sin ϕ 2 + ϕ 1 2 = δ P max 2 cos Δ ϕ 2 × sin k r 1 + r 2 2 ω t φ (0) 1 + φ (0) 2 2 ≡ δ P max 1+2 sin(const ω t ) where δ P max 1+2 = 2 δ P max cos Δ ϕ 2 = 2(0 . 0049 Pa) cos 9 . 84848 rad 2 = . 00206066 Pa is the pressure amplitude of the combined sound of the two speakers.sound of the two speakers....
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This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas.
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