481_PartUniversity Physics Solution

481_PartUniversity Physics Solution - 15-18 Chapter 15...

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15-18 Chapter 15 15.56. IDENTIFY: Apply 0 z τ = to one post and calculate the tension in the wire. / vF μ = for waves on the wire. vf λ = . The standing wave on the wire and the sound it produces have the same frequency. For standing waves on the wire, 2 n L n = . SET UP: For the 7 th overtone, 8 n = . The wire has / (0.732 kg)/(5.00 m) 0.146 kg/m mL = == . The free-body diagram for one of the posts is given in Figure 15.56. Forces at the pivot aren&t shown. We take the rotation axis to be at the pivot, so forces at the pivot produce no torque. EXECUTE: 0 z = gives cos57.0 ( sin57.0 ) 0 2 L wT L ⎛⎞ = ⎜⎟ ⎝⎠ °° . 235 N 76.3 N 2tan57.0 w T === . For waves on the wire, 76.3 N 22.9 m/s 0.146 kg/m F v = . For the 7 th overtone standing wave on the wire, 22 ( 5 . 0 0 m ) 1.25 m 88 L = . 22.9 m/s 18.3 Hz 1.25 m v f = . The sound waves have frequency 18.3 Hz and wavelength 344 m/s 18.8 m 18.3 Hz EVALUATE: The frequency of the sound wave is at the lower limit of audible frequencies. The wavelength of the standing wave on the wire is much less than the wavelength of the sound waves, because the speed of the waves on the wire is much less than the speed of sound in air. Figure 15.56 15.57. IDENTIFY: The magnitude of the transverse velocity is related to the slope of the t versus x curve. The transverse acceleration is related to the curvature of the graph, to the rate at which the slope is changing. SET UP: If y increases as t increases, y v is positive. y a has the same sign as y v if the transverse speed is increasing. EXECUTE: (a) and (b) (1): The curve appears to be horizontal, and 0 y v = . As the wave moves, the point will begin to move downward, and 0 y a < . (2): As the wave moves in the -direction, x + the particle will move upward so 0 y v > . The portion of the curve to the left of the point is steeper, so 0 y a > . (3) The point is moving down, and will increase its speed as the wave moves; 0 y v < , 0 y a < . (4) The curve appears to be horizontal, and 0 y v = . As the wave moves, the point will move away from the x -axis, and 0 y a > . (5) The point is moving downward, and will increase its speed as the wave moves; 0, < 0 yy va < . (6) The particle is moving upward, but the curve that represents the wave appears to have no curvature, so 0 and 0 >= . (c) The accelerations, which are related to the curvatures, will not change. The transverse velocities will all change sign. EVALUATE: At points 1, 3, and 5 the graph has negative curvature and 0 y a < . At points 2 and 4 the graph has positive curvature and 0 y a > . 15.58. IDENTIFY: The time it takes the wave to travel a given distance is determined by the wave speed v . A point on the string travels a distance 4 A in time T . SET UP: = . 1/ Tf = . EXECUTE: (a) The wave travels a horizontal distance d in a time () 8.00 m 0.333 s. 0.600 m 40.0 Hz dd t == = =
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Mechanical Waves 15-19 (b) A point on the string will travel a vertical distance of 4 A each cycle. Although the transverse velocity ( ) , y vx t is not constant, a distance of 8.00 m h = corresponds to a whole number of cycles, 3 (4 ) (8.00 m) [4(5.00 10 m)] 400, nh A == × = so the amount of time is (400) (40.0 Hz) 10.0 s.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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481_PartUniversity Physics Solution - 15-18 Chapter 15...

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