homework 35-solutions - klare (alk736) homework 35 Turner...

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Unformatted text preview: klare (alk736) homework 35 Turner (58220) 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 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 r 2 = 106 . 1 W 4 (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 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 P sin( ) , = k r- t- (0) . where r is the straight-line distance from the speaker to the listener. The two speakers have equal frequencies 1 = 2 = and equal initial phases (0) 1 = (0) 2 = (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 2- 1 = k r 2- t- (0)- k r 1- t- (0) = 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 = 2 f v = 2 (2820 Hz) 343 m / s = 51 . 6577 rad / m . Consequently, the phase di ff erence is = 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 P max 1 = P max 2 = P max = 0 . 0049 Pa at the listeners location. What is the pressure amplitude P max 1+2 of the combined sound of the two speakers? Correct answer: 0 . 00206066 Pa. Explanation: Let : P max = 0 . 0049 Pa . The two individual waves have pressures P 1 = P max sin( 1 ) , P 2 = P max sin( 2 ) . Adding them up to obtain the combined sound wave, we find P 1+2 = P 1 + P 2 = P max sin( 1 ) + sin( 2 ) = P max 2 cos 2- 1 2 sin 2 + 1 2 = P max 2 cos 2 sin k r 1 + r 2 2- t- (0) 1 + (0) 2 2 P max 1+2 sin(const- t ) where P max 1+2 = 2 P max cos 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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homework 35-solutions - klare (alk736) homework 35 Turner...

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