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Unformatted text preview: homework 25 – FIERRO, JEFFREY – Due: Apr 27 2008, 11:00 pm 1 Question 1, chap 16, sect 4. part 1 of 1 10 points A sphere having mass M = 1140 g is sup ported by a string that passes over a horizon tal rod of length L = 0 . 43 m (see the figure). The string makes an angle θ = 42 ◦ with the horizontal rod. The fundamental frequency is f = 26 Hz of standing waves in the string. The acceleration of gravity is 9 . 8 m / s 2 . L M θ Determine the mass of the string. Correct answer: 10 . 6713 g (tolerance ± 1 %). Explanation: The string length is L s = L cos θ = (0 . 43 m) cos(42 ◦ ) = 0 . 578622 m . The light rod would rotate if the string ex erted any vertical force on it. The rod exerts only a horizontal force on the string. For equilibrium of this bit of string summationdisplay F y = 0, or 0 = T sin θ − M g . Solving for T yields T = M g sin θ = (1 . 14 kg) (9 . 8 m / s 2 ) sin(42 ◦ ) = 16 . 6963 N . The wavelength of the fundamental mode is equal to twice the length of the rod λ = 2 L s = 2 (0 . 578622 m) = 1 . 15724 m . From the formula v = fλ = radicalBigg F μ = radicaltp radicalvertex radicalvertex radicalbt T m L s , where f is the fundamental frequency of the standing waves, and m is the mass of the string. We obtain m = T L s f 2 λ 2 = (16 . 6963 N) (0 . 578622 m) (26 Hz) 2 (1 . 15724 m) 2 = 1 . 14 kg = 10 . 6713 g . Question 2, chap 16, sect 4. part 1 of 1 10 points Two sound sources radiating in phase at a frequency of 430 Hz interfere such that max ima are heard at angles of 0 ◦ and 20 ◦ from a line perpendicular to that joining the two sources. The volicity of sound in air is 340 m/s. y L d S 1 S 2 θ listening direction θ δ Find the separation between the two sources if the velocity of sound is 340 m / s. Correct answer: 2 . 31185 m (tolerance ± 1 %). Explanation: Let : f = 430 Hz , v = 340 m / s , θ = 0 ◦ , and θ = 20 ◦ . homework 25 – FIERRO, JEFFREY – Due: Apr 27 2008, 11:00 pm 2 r 2 r 1 y L d S 1 S 2 θ = ta n − 1 parenleftBig y L parenrightBig listening direction δ ≈ d sin θ ≈ r 2 − r 1 P O negationslash S 2 Q S 1 ≈ 90 ◦ Q 2 π or 360 ◦ φ δ λ = 500 1 − 1 ◦ 90 ◦ 180 ◦ 270 ◦ 360 ◦ π/ 2 π 3 π/ 2 2 π φ 2 π = δ λ = d sin θ λ = d λ y radicalbig L 2 + y 2 ≈ d λ y L Because a maximum is heard at 0 ◦ and the sources are in phase, we can conclude that the path difference is 0. Because the next maximum is heard at 20 ◦ , the path difference to that position must be one wavelength....
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This note was uploaded on 05/11/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Mass, Work

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