(III) A cord stretched to a tension F T consists of two sections (as in Fig. 15–19) whose linear densities are in μ 1 and μ 2 . Take x = 0 to be the point (a knot) where they are joined, with μ 1 referring to that section of cord to the left and μ 2 that to the right. A sinusoidal wave, D = A sin[ k 1 ( x – v 1 t )], starts at the left end of the cord. When it reaches the knot, part of it is reflected and part is transmitted. Let the equation of the reflected wave be D R = A R sin[ k 1 ( x + v 1 t )] and that for the transmitted wave be D T = A T sin[ k 2 ( x – v 2 t )]. Since the frequency must be the same in both sections, we have ω 1 = ω 2 or k 1 v 1 = k 1 v 2 . ( a ) Because the cord is continuous, a point an infinitesimal distance to the left of the knot has the same displacement at any moment (due to incident plus reflected waves) as a point just to the right of the knot (due to the transmitted wave). Thus show that A = A T + A R . ( b ) Assuming that the slope (∂ D /∂ x ) of the cord just to the left of the knot is the same as the slope just to the right of the knot, show that the amplitude of the reflected wave is given by A R = ( υ 1 − υ 2 υ 1 + υ 2 ) A = ( k 2 − k 1 k 2 + k 1 ) A . ( c ) What is A T in terms of A ?
(III) A cord stretched to a tension F T consists of two sections (as in Fig. 15–19) whose linear densities are in μ 1 and μ 2 . Take x = 0 to be the point (a knot) where they are joined, with μ 1 referring to that section of cord to the left and μ 2 that to the right. A sinusoidal wave, D = A sin[ k 1 ( x – v 1 t )], starts at the left end of the cord. When it reaches the knot, part of it is reflected and part is transmitted. Let the equation of the reflected wave be D R = A R sin[ k 1 ( x + v 1 t )] and that for the transmitted wave be D T = A T sin[ k 2 ( x – v 2 t )]. Since the frequency must be the same in both sections, we have ω 1 = ω 2 or k 1 v 1 = k 1 v 2 . ( a ) Because the cord is continuous, a point an infinitesimal distance to the left of the knot has the same displacement at any moment (due to incident plus reflected waves) as a point just to the right of the knot (due to the transmitted wave). Thus show that A = A T + A R . ( b ) Assuming that the slope (∂ D /∂ x ) of the cord just to the left of the knot is the same as the slope just to the right of the knot, show that the amplitude of the reflected wave is given by A R = ( υ 1 − υ 2 υ 1 + υ 2 ) A = ( k 2 − k 1 k 2 + k 1 ) A . ( c ) What is A T in terms of A ?
(III) A cord stretched to a tension FT consists of two sections (as in Fig. 15–19) whose linear densities are in μ1 and μ2. Take x = 0 to be the point (a knot) where they are joined, with μ1 referring to that section of cord to the left and μ2 that to the right. A sinusoidal wave, D = A sin[k1(x – v1t)], starts at the left end of the cord. When it reaches the knot, part of it is reflected and part is transmitted. Let the equation of the reflected wave be DR = AR sin[k1(x + v1t)] and that for the transmitted wave be DT = AT sin[k2(x – v2t)]. Since the frequency must be the same in both sections, we have ω1 = ω2 or k1v1 = k1v2. (a) Because the cord is continuous, a point an infinitesimal distance to the left of the knot has the same displacement at any moment (due to incident plus reflected waves) as a point just to the right of the knot (due to the transmitted wave). Thus show that A = AT + AR. (b) Assuming that the slope (∂D/∂x) of the cord just to the left of the knot is the same as the slope just to the right of the knot, show that the amplitude of the reflected wave is given by
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In Fig. 16-42, a string, tied to a sinusoidal oscillator at Pand running over a support at Q, is stretched by a block of mass m.The separation L between P and Q is 1.20 m, and the frequency fof the oscillator is fixed at 120 Hz.The amplitude of the motion atP is small enough for that point to be considered a node.A nodealso exists at Q. A standing wave appears when the mass of thehanging block is 286.1 g or 447.0 g, but not for any intermediatemass.What is the linear density of the string?
A standard trumpet has a resonant tube length of 1.4 m and is closed at one end. What are the first three resonant frequencies if the speed of sound is 340 m/s?
Calculate the lowest resonant frequency for a brick partition 150 mm thick, 6 m by 3 m in an area with a longitudinal wave velocity of 2350 m/s. (Assume it is supported at its edges.)
Chapter 15 Solutions
Physics for Scientists and Engineers, Vol 1 (Chapters 1-20)
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