hw14 - al-ameri (aoa434) – Hwk14 – Stokes –...

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Unformatted text preview: al-ameri (aoa434) – Hwk14 – Stokes – (19102)1This print-out should have 22 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This is ONLINE HOMEWORK No 14. Itis due before 6.00 am on Wednesday 13 May,Abu Dhabi Time (9.00 pm Tuesday in Texas)00110.0 pointsIf the frequency of sound is doubled, how willits speed change? How will its wavelengthchange?1.Its speed will double, and its wavelengthwill not change.2.There will be no change in its speed andwavelength.3.Its speed will not change, and its wave-length will halve.correct4.Its speed will halve, and its wavelengthwill double.5.Both speed and wavelength will double.Explanation:v=fλ= (2f)parenleftbigg12λparenrightbiggThe speed of sound is constant in a givenmedium. If the frequency of sound is doubled,its speed will not change, but its wavelengthmust be “compressed” to half. The speed ofsound depends only on the medium throughwhich it travels, not on its frequency, wave-length, or intensity (until the intensity is sogreat that a shock wave results.)002(part 1 of 2) 10.0 pointsA stone is dropped into a deep canyon and isheard to strike the bottom 12.3 s after release.The speed of sound waves in air is 343 m/sand the acceleration of gravity is 9.8 m/s2.How deep is the canyon?Correct answer: 558.147 m.Explanation:Let :t= 12.3 s,vs= 343 m/s,andg= 9.8 m/s2.Letdbe the distance the stone drops. Thenthe time for the stone to do drop and then forus to hear the sound ist=tgravity+tsound=radicalBigg2dg+dvst−dvs=radicalBigg2dgvst−d=vsradicalBigg2dg.Squaring both sides,v2st2−2vst d+d2=2v2sdgd2−parenleftbigg2vst+2v2sgparenrightbiggd+ (vst)2= 0.This equation can be written asa d2+b d+c= 0,so thatd=−b±√b2−4a c2awherea= 1,b=−2vst−2v2sg=−2 (343 m/s)(12.3 s)−2 (343 m/s)2(9.8 m/s2)=−32447.8 m,and the discriminant isD=b2−4a c=b2−4 (vst)2= (−32447.8 m)2−4 [(343 m/s)(12.3 s)]2= 9.81663×108m2.al-ameri (aoa434) – Hwk14 – Stokes – (19102)2Thus the stone will fall a distanced=−−32447.8 m2+√9.81663×108m22= 558.147 m.003(part 2 of 2) 10.0 pointsWhat would be the percentage error in thedepth if the time required for the sound toreach the canyon rim were ignored?Correct answer: 32.8182%.Explanation:If we ignore the speed of sound, then thedepth would bed′=12g t2=12(9.8 m/s2) (12.3 s)2= 741.321 m,and the percentage error isError =d′−dd×100%=741.321 m−558.147 m558.147 m×(100%)= 32.8182%.004(part 1 of 4) 10.0 pointsAs a sound wave travels through the air, itproduces pressure variations (above and be-low atmospheric pressure) given byP= (1.27 Pa) sinbracketleftbig(πm−1)x−(340πs−1)tbracketrightbigwherexis in meters andtis in seconds....
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hw14 - al-ameri (aoa434) – Hwk14 – Stokes –...

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