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# Chapter-17 - Chapter 17 1(a When the speed is constant we...

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735 Chapter 17 1. (a) When the speed is constant, we have v = d / t where v = 343 m/s is assumed. Therefore, with t = 15/2 s being the time for sound to travel to the far wall we obtain d = (343 m/s) × (15/2 s) which yields a distance of 2.6 km. (b) Just as the 1 2 factor in part (a) was 1/( n + 1) for n = 1 reflection, so also can we write ( 29 ( ( 343 15 15s 343m/s 1 1 d n n d = = - + for multiple reflections (with d in meters). For d = 25.7 m, we find n = 199 2 2.0 10 × . 2. The time it takes for a soldier in the rear end of the column to switch from the left to the right foot to stride forward is t = 1 min/120 = 1/120 min = 0.50 s. This is also the time for the sound of the music to reach from the musicians (who are in the front) to the rear end of the column. Thus the length of the column is 2 (343m/s)(0.50s) =1.7 10 m. l vt = = × 3. (a) The time for the sound to travel from the kicker to a spectator is given by d / v , where d is the distance and v is the speed of sound. The time for light to travel the same distance is given by d / c , where c is the speed of light. The delay between seeing and hearing the kick is t = ( d / v ) – ( d / c ). The speed of light is so much greater than the speed of sound that the delay can be approximated by t = d / v . This means d = v t . The distance from the kicker to spectator A is d A = v t A = (343 m/s)(0.23 s) = 79 m. (b) The distance from the kicker to spectator B is d B = v t B = (343 m/s)(0.12 s) = 41 m. (c) Lines from the kicker to each spectator and from one spectator to the other form a right triangle with the line joining the spectators as the hypotenuse, so the distance between the spectators is ( 29 ( 29 2 2 2 2 79m 41m 89m A B D d d = + = + = . 4. The density of oxygen gas is

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CHAPTER 17 736 3 3 0.0320kg 1.43kg/m . 0.0224m = = ρ From / v B ρ = we find ( 29 ( 2 2 3 5 317m/s 1.43kg/m 1.44 10 Pa. B v = = = × ρ 5. Let t f be the time for the stone to fall to the water and t s be the time for the sound of the splash to travel from the water to the top of the well. Then, the total time elapsed from dropping the stone to hearing the splash is t = t f + t s . If d is the depth of the well, then the kinematics of free fall gives 2 1 2 f d gt = 2 / . f t d g = The sound travels at a constant speed v s , so d = v s t s , or t s = d / v s . Thus the total time is 2 / / s t d g d v = + . This equation is to be solved for d . Rewrite it as 2 / / s d g t d v = - and square both sides to obtain 2 d / g = t 2 – 2( t / v s ) d + (1 + 2 s v ) d 2 . Now multiply by g 2 s v and rearrange to get gd 2 – 2 v s ( gt + v s ) d + g 2 s v t 2 = 0. This is a quadratic equation for d . Its solutions are ( 29 2 2 2 2 2 2 ( ) 4 4 . 2 s s s s s v gt v v gt v g v t d g + ± + - = The physical solution must yield d = 0 for t = 0, so we take the solution with the negative sign in front of the square root. Once values are substituted the result d = 40.7 m is obtained. 6. Using Eqs. 16-13 and 17-3, the speed of sound can be expressed as B v f λ ρ = = , where ( / ) / B dp dV V = - . Since , and V λ ρ are not changed appreciably, the frequency ratio becomes
737 ( / ) ( / ) s s s s i i i i f v B dp dV f v B dp dV = = = .

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