final_oldfinal1

# final_oldfinal1 - oldnal 01 PAPAGEORGE, MATT Due: May 6...

This preview shows pages 1–3. Sign up to view the full content.

oldfnal 01 – PAPAGEORGE, MATT – Due: May 6 2008, 4:00 am 1 Question 1, chap 16, sect 3. part 1 oF 1 10 points While waiting For Stan Speedy to arrive on a late passenger train, Kathy Kool notices beats occurring as a result oF two trains blow- ing their whistles simultaneously. One train is at rest and the other is approaching her at a speed oF 13 . 9 km / h. Assume that both whistles have the same Frequency and that the speed oF sound is 344 m / s. IF Kathy hears beat Frequency at 7 . 39 Hz, what is the Frequency oF the whistles? Correct answer: 651 . 011 Hz (tolerance ± 1 %). Explanation: The Frequency oF sound waves produced by the moving whistle at Kathy’s location is given, according to Doppler E±ect, by f ± = ± v sound v sound - v train ² f = ± 344 m / s 344 m / s - 3 . 86111 m / s ² f = (1 . 01135) f. The beat Frequency equals the di±erence in Frequency between the two sources. f b = | f ± - f | = (1 . 01135 - 1) ThereFore, f = f b (1 . 01135 - 1) = 7 . 39 Hz (1 . 01135 - 1) = 651 . 011 Hz . Question 2, chap 16, sect 4. part 1 oF 1 10 points An elastic string oF mass 5 . 2 g is stretched to a length oF 1 . 791 m by the tension Force 44 N. The string is fxed at both ends and has Fundamental Frequency f 1 . When the tension Force increases to 624 . 8 N the string stretches to a length 8 . 68635 m and its Fundamental Frequency becomes f 2 . Calculate the ratio f 2 f 1 . Correct answer: 1 . 71109 (tolerance ± 1 %). Explanation: Let : L 1 =1 . 791 m , F 1 = 44 N , L 2 =8 . 68635 m , and F 2 = 624 . 8N . The Fundamental mode oF a string has wavelength λ =2 L and thereFore Frequency f = v 2 L , where v = ³ F μ = ´ µ µ F m L = · FL m is the speed oF waves on the string. Alto- gether, f = 1 2 L · m = 1 2 · F mL . Thus the ratio is f 2 f 1 = 1 2 · F 2 2 1 2 · F 1 1 = · F 2 L 1 F 1 L 2 = ³ (624 . 8 N) (1 . 791 m) (44 N) (8 . 68635 m) = 1 . 71109 . Note: You don’t need the value oF the string’s mass since it cancels out oF the Frequency ratio. Question 3, chap 17, sect 3. part 1 oF 1 10 points A variable-length air column is placed just below a vibrating wire that is fxed at the both ends. The length oF air column open at

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
oldfnal 01 – PAPAGEORGE, MATT – Due: May 6 2008, 4:00 am 2 one end is gradually increased From zero until the frst position oF resonance is observed at 24 . 9 cm. The wire is 127 . 2 cm long and is vibrating in its third harmonic. ±ind the speed oF transverse waves in the wire iF the speed oF sound in air is 340 m / s. Correct answer: 289 . 478 m / s (tolerance ± 1 %). Explanation: Given : L = 24 . 9 cm , L wire = 127 . 2 cm , and v sound = 340 m / s . The length oF the air column when the frst resonance is heard is L = λ air 4 , where λ air is the wavelength oF the sound in air. The Frequency oF sound wave, and hence the vibrating wire producing sound, is f = v sound λ air = v sound 4 L . When the wire vibrates in its third harmonic, its wavelength is related to its length as Fol- lows; L wire = 3 2 λ wire , where λ wire is the wavelength oF waves trav- eling in the wire. ThereFore, λ wire = 2 3 L wire , and the speed oF transverse waves in the wire is v wire = λ wire f = ± 2 3 L wire ² ³ v sound 4 L ´ = L wire v sound 6 L = (127 . 2 cm) (340 m / s) 6 (24 . 9 cm) = 289 . 478 m / s .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas at Austin.

### Page1 / 23

final_oldfinal1 - oldnal 01 PAPAGEORGE, MATT Due: May 6...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online