OLDHW35 - oldhomewk 35 PAPAGEORGE MATT Due 4:00 am Question...

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oldhomewk 35 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 1 Question 1, chap 16, sect 4. part 1 of 1 10 points Two guitar strings, of equal length and lin- ear density, are tuned such that the second harmonic of the first string has the same fre- quency as the third harmonic of the second string. The tension of the first string is 210 N. Calculate the tension of the second string. Correct answer: 93 . 3333 N (tolerance ± 1 %). Explanation: The frequency of the n th harmonic of a string is f = n 2 l radicalBigg F μ . (This arises from the requirement that an integer number of half-wavelengths must fill the string in order to keep the end-points of the string fixed.) For two strings of equal lengths and linear densities, we get f 2 f 1 = n 2 n 1 radicalbigg F 2 F 1 . In this case f 2 = f 1 , so F 2 = F 1 parenleftbigg n 1 n 2 parenrightbigg 2 = 93 . 3333 N . Question 2, chap 16, sect 4. part 1 of 1 10 points Two sound sources radiating in phase at a frequency of 540 Hz interfere such that max- ima are heard at angles of 0 and 26 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: 1 . 43629 m (tolerance ± 1 %). Explanation: Let : f = 540 Hz , v = 340 m / s , θ 0 = 0 , and θ = 26 . r 2 r 1 y L d S 1 S 2 θ = tan 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 0 1 - 1 0 90 180 270 360 0 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 26 , the path difference to that position must be one wavelength.
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oldhomewk 35 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 2 sin θ = Δ s d d = λ sin θ = v f sin θ = 340 m / s (540 Hz) sin 26 = 1 . 43629 m . For additional intensity maxima, d sin θ m = mλ, m = 2 , 3 , 4 , ... θ m = sin 1 parenleftbigg d parenrightbigg , m = 2 , 3 , 4 , ... The next angle is for m = 2: θ m =2 = sin 1 parenleftbigg d parenrightbigg = sin 1 parenleftbigg m v f d parenrightbigg = sin 1 bracketleftbigg 2 (340 m / s) (540 Hz) (1 . 43629 m) bracketrightbigg . Question 3, chap 16, sect 4. part 1 of 1 10 points A sinusoidal wave in a rope is described by the wave function y = A sin( k x + ω t ) , where x and y are in meters, t is in seconds, A = 0 . 12 m, k = 0 . 68 m 1 , and ω = 24 Hz. The acceleration of gravity is 9 . 8 m / s 2 and the rope has a linear mass density of 1 . 1 g / m. vibrator 24 Hz μ = 1 . 1 g / m m 23 . 0999 m If the tension in the rope is provided by an arrangement like the one illustrated above, what is the value of the suspended mass? Correct answer: 0 . 139821 kg (tolerance ± 1 %).
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