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Mays Overton, Homework 34 Due: Apr 29 2005, 4:00 am Inst: Turner This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Given: At this altitude, the speed of sound in air at 273 C is 332 m/s. A group of hikers hear an echo 3.9 s after they shout. The temperature is 23 C. How far away is the mountain that reflected the sound wave? Correct answer: 674.12 m. Explanation: (23 C) (273 C) Thus, the time lapse is t = t2 - t1 1 1 =d - v2 v1 = (30 km) 1 1 - 333 m/s 3 108 m/s = 90.09 s . 1 v = (332 m/s) 1+ = 345.703 m/s . If d is the distance to the object, the total distance traveled by the sound (there and back) is 2 d . Thus, d=v t 2 (3.9 s) 2 = (345.703 m/s) = 674.12 m . 003 (part 1 of 1) 10 points Carol drops a stone into a mine shaft 143.7 m deep. Take the speed of sound as 343 m/s. The acceleration of gravity is 9.8 m/s2 . How soon after she drops the stone does she hear it hit the bottom of the shaft? Correct answer: 5.83435 s. Explanation: The speed of sound in air at room temperature is vs = 343 m/s. Two times are involved: 1) The stone travels to the bottom via free fall: 1 d = g t2 , 2 1 so that 2d t1 = g 2) The sound travels back to the top of the shaft: d t2 = . vs Thus the total time is t = t 1 + t2 = = d 2d + g vs 002 (part 1 of 1) 10 points Given: The speed of the lightwaves in air is 3 108 m/s. The speed of sound waves in air is 333 m/s. What is the time lapse between seeing a lightening strike and hearing the thunder if the lightening flash is 30 km away? Correct answer: 90.09 s. Explanation: The time for the light to travel 30 km is t1 = d . v1 2 (143.7 m) (143.7 m) + (9.8 m/s2 ) (343 m/s) = 5.83435 s . 004 (part 1 of 1) 10 points Consider a longitudinal (compressional) wave of wavelength 1.08 m traveling with speed The time for the sound to travel 30 km is d . t2 = v2 Overton, Mays Homework 34 Due: Apr 29 2005, 4:00 am Inst: Turner 4810 m/s along the x-direction through a medium of density 8290 kg/m3 . The maximum displacement of the molecules in the medium from their equilibrium position is 2.97 10-10 m. Calculate the maximum pressure variation in this medium. Correct answer: 331.404 Pa. Explanation: Using Pm = v sm , 2v , we obtain for the maximum pressure variation = Pm = 2 v 2 sm 2 (8290 kg/m3 ) = (1.08 m) (4810 m/s)2 (2.97 10-10 m) = 331.404 Pa . and 2 006 (part 1 of 1) 10 points The noise level coming from a pig pen with 239 pigs is 78.5 dB. Assuming each of the remaining pigs squeal at their original level after 75 of their companions have been slaughtered, what is the decibel level of the remaining pigs? Correct answer: 76.8645 dB. Explanation: The intensity of a sound wave with sound level we calculate using: I = I0 10/10 . For two sound levels, the intensities are respectively: I1 = I0 101 /10 , I2 = I0 102 /10 . Therefore, 2 = 1 + 10 log10 = 1 + 10 log10 I2 I1 N2 N1 239 - 75 239 005 (part 1 of 1) 10 points A microwave oven generates a sound level of 46.4 dB when consuming 0.554 kW of power. Estimate the fraction of this power that is converted into the energy of sound waves. Assume the dimensions of the oven are 21.3 cm by 41.1 cm by 70.7 cm and assume the sound level is uniform over the outer surface. Correct answer: 8.3318 10-11 . Explanation: The outer surface of the microwave has area A = 2 (L w + L h + h w) = 1.05742 m2 . The intensity of the sound is given by The sound power is I = I0 10/10 = 4.36516 10-8 W/m2 . Psound = I A = 4.61582 10-8 W . = 78.5 dB + 10 log10 = 76.8645 dB. Then the oven's efficiency as a sound source is Psound Efficiency = P 4.61582 10-8 W = (0.554 kW)(1000 W/kW) = 8.3318 10-11 . 007 (part 1 of 1) 10 points two Consider loud bells: The sound intensity of the first bell is 1 = 93 db (loud!) and the sound intensity of the second bell 2 = 106 db (even louder!). Assume each bell produces a harmonic sound wave or respective pressure amplitudes max max P1 and P2 , and calculate the ratio max P2 max P1 of the two pressure amplitudes. Overton, Mays Homework 34 Due: Apr 29 2005, 4:00 am Inst: Turner Correct answer: 4.46684 . Explanation: First. let us compare the absolute sound intensities of the two bells. The decibel measure of the sound volume is logarithmic: = 10 log10 hence I2 102 /10 = I1 101 /10 = 10(2 -1 )/10 = 10(106 db-93 db)/10 = 1013 db/10 = 19.9526. Now consider the wave amplitudes. A harmonic sound wave of displacement amplitude smax has pressure amplitude P max = smax v and intensity I = 1 2 v (smax )2 . 2 (P max )2 2 v I = I = I0 10/10 , I0 3 A stone is thrown into a quiet pool of water and produces a water wave spreading spherically. With no fluid friction, the amplitude of the wave falls off with distance r from the impact point as 1 1. correct r 1 2. 3 r 1 3. 3/2 r 1 4. r 1 5. 2 r Explanation: The intensity of the two dimensional wave is proportional to the inverse of the distance and its amplitude is proportional to the square root of the intensity. Therefore its amplitude is proportional to the square root of the distance. 009 (part 1 of 2) 10 points A skyrocket explodes 108 m above the ground. Three observers are spaced 87 m apart, with observer A directly under the point of the explosion. Combining the two formulas, we find I = regardless of the sound frequency. Consequently, for two sound waves in the same air I2 = I1 and conversely max P2 max P1 max P2 max P1 2 108 m , A 87 m B 87 m C = I2 = 10(2 -1 )/20 . I1 For the two bells is question, we have max P2 = 19.9526 = 4.46684. max P1 Find the ratio of the sound intensity heard by observer A to that heard by observer B. Correct answer: 1.64892 . Explanation: 008 (part 1 of 1) 10 points Given : h = 108 m and d = 87 m . Overton, Mays Homework 34 Due: Apr 29 2005, 4:00 am Inst: Turner Source 4 rC rB B Assume: The receiver has a cross section A perpendicular to the direction of the incident sound. What is the intensity of the sound waves detected by a receiver a distance r from the source? 1. I = 4 r 2 W 2. I = r 2 W 3. I = W 2rr C 87 m 87 m The intensity at a distance r from the source is P I= , 4 r2 and distances from the source to points A and B are rA = h and rA = 108 m A 4. I = r 2 W 5. I = 2 r W 6. I = W r2 W 7. I = r2 W 8. I = correct 4 r2 W 9. I = A rB = so h 2 + d2 , r2 IA h2 + d2 = B = 2 IB h2 rA (108 m)2 + (87 m)2 = (108 m)2 = 1.64892 . 010 (part 2 of 2) 10 points Find the ratio of the intensity heard by observer A to that heard by observer C. Correct answer: 3.59568 . Explanation: The distance from the source to C is rC = so IA r2 h2 + 4 d 2 = C = 2 IC h2 rA (108 m)2 + 4 (87 m)2 = (108 m)2 = 3.59568 . 011 (part 1 of 2) 10 points A point source emits sound waves uniformly in all directions with a power output W . rock concert r A Explanation: Basic Concept: = 10 log I I0 h2 + (2 d)2 , Solution: The intensity of the sound waves is given by W I= . 4 r2 012 (part 2 of 2) 10 points The average sound level of a typical musical instrument in a musical group is dB. What is the sound level in dB if 10 instruments are simultaneously playing? 1. = + 20 2. = 10 3. = + 1 Overton, Mays Homework 34 Due: Apr 29 2005, 4:00 am Inst: Turner 4. = 5. = 2 6. = 20 7. = + 10 correct 8. = 5 9. = + 2 10. = + 5 Explanation: The intensity of the sound waves generated by a single instrument is I = I0 10/10 . The intensity generated by 10 such instruments is I = 10 I = I0 10[1+(/10)] , so = 10 log I I0 5 = 10 + . ... View Full Document

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