ch16af - Chapter 16 Superposition and Standing Waves 86...

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Chapter 16 Superposition and Standing Waves 86 ·· Three successive resonance frequencies in an organ pipe are 1310, 1834, and 2358 Hz. ( a ) Is the pipe closed at one end or open at both ends? ( b ) What is the fundamental frequency? ( c ) What is the length of the pipe? ( a ) If the pipe is open at both ends then f = f 1 , and f n = nf 1 , where n is an integer. f = 524 Hz and 1310 Hz = 2.5 f 1 . So the pipe is closed at one end. ( b ) For a pipe closed at one end, f = 2 f 1 , f 1 = 262 Hz. ( c ) L = λ /4; λ = 340/262 m = 1.30 m. L = 32.4 cm. 87 ·· A wire of mass 1 g and length 50 cm is stretched with a tension of 440 N. It is then placed near the open end of the tube in Example 16-8 and stroked with a violin bow so that it oscillates at its fundamental frequency. The water level in the tube is then lowered until a resonance is obtained, which occurs at 18 cm below the top of the tube. Use the data given to determine the speed of sound in air. Why is this method not very accurate? 1. Lm F/ = L v/ = f 4 2 1 3 1 10 2 / 440 - × = f Hz = 469 Hz 2. v s = f λ = 4 fL v s = 338 m/s The method is not very accurate because it neglects end effects. (see Problem 51) 88 ·· On a windy day, a drain pipe will sometimes resonate. Estimate the resonance frequency of a drain pipe on a single-story house. How much might this frequency change from winter to summer in your region? Pipe length 5 m. Pipe open at both ends, so λ 1 10 m, and f v / λ = 34 Hz. The frequency will be somewhat higher in the summer because v T . 89* ·· A 50-cm-long wire fixed at both ends vibrates with a fundamental frequency f 0 when the tension is 50 N. If the tension is increased to 60 N, the fundamental frequency increases by 5 Hz, and a further increase in tension to 70 N results in a fundamental frequency of ( f 0 + 7) Hz. Determine the mass of the wire. 1. f F 2. m = F /4 Lf 0 2 (see Problem 87) 0 0 0 ; 0954 . 1 50 / 60 / ) 5 ( f f f = = + = 52.4 Hz m = 50/(4 × 0.5 × 52.4 2 ) kg = 9.1 g 90 ·· A standing wave on a rope is represented by the following wave function: t x t x y π π 40 cos 2 sin 02 . 0 ) , ( = , where x and y are in meters and t is in seconds. ( a ) Write wave functions for two traveling waves that when superimposed will produce the resultant standing-wave pattern. ( b ) What is the distance between the nodes of the standing wave? ( c ) What is the velocity of a segment of the rope at x = 1 m? ( d ) What is the acceleration of a segment of the rope at x = 1 m? ( a ) y 1 ( x, t ) = 0.01 sin [( π x /2) - 40 π t ] m; y 2 ( x, t ) = 0.01 sin [( π x /2) + 40 π t ] m. ( b ) Distance between nodes = λ /2; λ = 2 π / k = 4 m; distance between nodes = 2 m. ( c ) x = 1 m is an antinode. v y (1, t ) = - 0.8 π sin (40 π t ) m/s = - 2.51 sin (40 π t ) m/s.
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Chapter 16 Superposition and Standing Waves ( d ) a y (1, t ) = - 316 cos(40 π t ) m/s 2 .
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