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(5%) Problem 16: A crude approximation of human voice production is to consider the breathing passages and mouth to be a resonating tube closed at
(5%) Problem 16: A crude approximation of human voice production is to consider the breathing passages and mouth to be a resonating tube closed at one end. (See the figure) - Nasal cavity Soft palate . Tongue - Epiglottis Pharynx Vocal cord Larynx Trachea Otheexpertta.com @theexpertta.com - tracking id: 9N69-4E-C3-45-84AD-31944. In accordance with Expert TA's Terms of Service. copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account. 50% Part (a) What is the fundamental frequency, in hertz, of a voice if the tube is 0.255-m long, taking air temperature to be 37.0*C? f1 = 345 Correct! *50% Part (b) What would this frequency become, in hertz, if the person replaced the air with helium? Assume the same temperature dependence for the speed of sound in helium as for air, and that the speed of sound in Helium is 965 m/s at 0-C. Grade Summary $2 = Deductions 0% Potential 100%(5%) Problem 17: You have been hired by a professional musician as an interior designer. Your task is to design a very special enclosed rectangular glass shower stall. The specifications are based on which notes, when sung in the shower, will resonate. Your 1.9 meter tall client takes showers so that the air temperature is That = 32.2'C . Your client wants the shower stall to resonate precisely at f, = 150 Hz, f2 = 192 Hz, and f3 = 418 Hz. Note that the stall is to be wider than it is deep, and that no dimension of the shower stall can be less than 50 cm. Assume that each dimension of the shower works like a resonance tube that is closed on both ends. @theexpertta.com - tracking id: 9N69-4E-C3-45-84AD-31944. In accordance with Expert TA's Terms of Service. copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account. A 20% Part (a) Calculate the speed of sound in air, vs (in m/s), while your client is showering. Recall that the speed of sound in air (in m/s) can be found using vs = 331.5 v(1 + [7/273.15]), where 7 is the air temperature in C. A 20% Part (b) How tall (in meters) should the shower stall be if it is to resonate at the lowest frequency and your client can stand up fully while showering? A 20% Part (c) Calculate the minimum depth (in meters) of the shower stall. & 20% Part (d) Calculate the minimum width (in meters) of the shower stall. A 20% Part (e) At these frequencies, trained musicians can generally distinguish 2 notes whose ratio is larger than 1.00579 (10 cents). If your client decides to take a shower that is 10 C colder than that specified, what is the ratio of the hot to the cold resonant frequencies? Grade Summary ful fc = Deductions 0% 100 0(5%) Problem 21: Musicians need to be able to discern frequencies which are quite near each other. Assume that the average musician can differentiate between frequencies that vary by only 0.6%. This corresponds to about 1/10 of the frequency difference between neighboring notes in the middle of the piano keyboard. @theexpertta.com - tracking id: 9N69-4E-C3-45-84AD-31944. In accordance with Expert TA's Terms of Service. copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account. A 50% Part (a) What is the closest lower frequency, in hertz, to 465 Hz that a musician can clearly distinguish as being different in frequency from 465 Hz? & 50% Part (b) What is the closest higher frequency, in hertz, to 465 Hz that a musician can clearly distinguish as being different in frequency from 465 Hz? Grade Summary f+ = Deductions 0%
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