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homework 35

# homework 35 - ramos(dar2447 homework 35 Turner(58120 This...

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ramos (dar2447) – homework 35 – Turner – (58120) 1 This print-out should have 14 questions. Multiple-choice 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 94.71 W. P x 8 . 94 m What is the intensity of the sound waves at point x , 8.94 m away? Correct answer: 0 . 0942998 W / m 2 . Explanation: Let : P = 94 . 71 W and r = 8 . 94 m . I = P 4 π r 2 = 94 . 71 W 4 π (8 . 94 m) 2 = 0 . 0942998 W / m 2 . 002 (part 1 of 2) 10.0 points Two loudspeakers are placed on a wall 2 . 53 m apart. A listener stands directly in front of one of the speakers, 46 . 1 m from the wall. The speakers are being driven by the same electric signal generated by a harmonic oscillator of frequency 5760 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: 7 . 31967 rad. Explanation: Let : r 1 = 46 . 1 m , f = 5760 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 straight-line 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 = 46 . 1 m but the distance to the second speaker is r 2 = radicalBig (46 . 1 m) 2 + (2 . 53 m) 2 = 46 . 1694 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 wave-number 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 (5760 Hz) 343 m / s = 105 . 514 rad / m . Consequently, the phase difference is Δ ϕ = k ( r 2 r 1 ) = (105 . 514 rad / m) × (46 . 1694 m 46 . 1 m) = 7 . 31967 rad .

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ramos (dar2447) – homework 35 – Turner – (58120) 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 . 0043 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 . 00747074 Pa. Explanation: Let : δP max = 0 . 0043 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 φ (0) 1 + φ (0) 2 2 δP max 1+2 sin(const ω t ) where δP max 1+2 = 2 δP max vextendsingle vextendsingle vextendsingle vextendsingle cos parenleftbigg Δ ϕ 2 parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle = 2(0 . 0043 Pa) vextendsingle vextendsingle vextendsingle vextendsingle cos parenleftbigg 7 . 31967 rad 2 parenrightbiggvextendsingle vextendsingle vextendsingle
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