<p>P18.3 (a) y1 = f( x - vt) , so wave 1 travels in the +x direction</p><p> y2 = f( x + vt) , so wave 2 travels in the -x direction</p><p>5+ 5 (b) To cancel, y+ y = 0 : = 1 2 (3x- 4 t)2 + 2( 3 x + 4 t - 6) 2 + 2</p><p>(3x- 4 t)2 =( 3 x + 4 t - 6) 2 3x- 4 t =�( 3 x - 4 t 6)</p><p> for the positive root, 8t = 6 t = 0.750 s</p><p>(at t = 0.750 s , the waves cancel everywhere)</p><p>(c) for the negative root, 6x = 6 x = 1.00 m</p><p>(at x = 1.00 m , the waves cancel always)</p><p>P18.5 The resultant wave function has the form 骣f 骣 f y=2 A cos琪 sin 琪 kx -w t + 0 桫2 桫 2</p><p>骣f 轾-p 4 (a) A=2 A0 cos琪 = 2( 5.00) cos犏 = 9.24 m 桫2臌 2</p><p> w 1 200p (b) f = = = 600 Hz 2p 2 p P18.8 (a) Dx =9.00 + 4.00 - 3.00 = 13 - 3.00 = 0.606 m v 343 m s The wavelength is l = = = 1.14 m f 300 Hz</p><p>Dx 0.606 Thus, = = 0.530 of a wave , l 1.14</p><p> or Df =2 p ( 0.530) = 3.33 rad</p><p>Dx D x (b) For destructive interference, we want =0.500 = f l v</p><p> v 343 where Dx is a constant in this set up. f = = = 283 Hz 2Dx 2( 0.606)</p><p>P18.15 The facing speakers produce a standing wave in the space between them, with the spacing between nodes being</p><p> l v 343 m s dNN = = = = 0.214 m 2 2 f 2( 800 s-1 )</p><p>If the speakers vibrate in phase, the point halfway between them is an antinode of pressure at a distance from either speaker of</p><p>1.25 m = 0.625 . 2</p><p>0.214 Then there is a node at 0.625- = 0.518 m 2</p><p> a node at 0.518 m- 0.214 m = 0.303 m</p><p> a node at 0.303 m- 0.214 m = 0.089 1 m</p><p> a node at 0.518 m+ 0.214 m = 0.732 m</p><p> a node at 0.732 m+ 0.214 m = 0.947 m</p><p> and a node at 0.947 m+ 0.214 m = 1.16 m from either speaker. v P18.19 L = 30.0 m ; m = 9.00 10-3 kg m ; T = 20.0 N ; f = 1 2L</p><p>1 2 骣T where v = = 47.1 m s 桫琪m</p><p>47.1 so f = = 0.786 Hz f2=2 f 1 = 1.57 Hz 1 60.0</p><p> f3=3 f 1 = 2.36 Hz f4=4 f 1 = 3.14 Hz</p><p> l EMBED MSDraw \* mergeformat P18.24 For the whole string vibrating, d =0.64 m = ; l = 1.28 m . The NN 2</p><p>1 speed of a pulse on the string is v= f l =330 1.28 m = 422 m s . s</p><p>2 l (a) When the string is stopped at the fret, d =0.64 m = ; NN 3 2 l = 0.853 m</p><p> v 422 m s FIG. P18.24(a) f = = = 495 Hz . l 0.853 m</p><p>(b) The light touch at a point one third of the way along the string EMBED MSDraw \* mergeformat damps out vibration in the two lowest vibration states of the string as a whole. The whole string vibrates in its third resonance</p><p> l possibility: 3d = 0.64 m = 3 ; l = 0.427 m NN 2</p><p> v 422 m s FIG. P18.24(b) f = = = 990 Hz . l 0.427 m *P18.39 Assuming an air temperature of T = 37� C 310 K , the speed of sound inside the pipe is</p><p>310 K v =(331 m s) = 353 m s . 273 K</p><p>In the fundamental resonant mode, the wavelength of sound waves in a pipe closed at one end is l = 4L . Thus, for the whooping crane</p><p>1 v (353 m s) 骣3.281 ft l =4( 5.0 ft) = 2.0 10 ft and f = =琪 = 57.9 Hz . l 2.0 101 ft 桫 1 m</p><p>P18.41 For a closed box, the resonant frequencies will have nodes at both sides, so the permitted wavelengths </p><p>1 will be L= nl , (n = 1, 2, 3, ) . 2</p><p> nl nv nv i.e., L = = and f = . 2 2 f 2L Therefore, with L = 0.860 m and L = 2.10 m , the resonant frequencies are</p><p> fn = n (206 Hz) for L = 0.860 m for each n from 1 to 9</p><p> and fn = n (84.5 Hz) for L = 2.10 m for each n from 2 to 23.</p><p>540 P18.51 f档 v T f =110 = 104.4 Hz new 600</p><p>Df = 5.64 beats s l 1.75 m P18.62 The second standing wave mode of the air in the pipe reads ANAN, with d = = NA 4 3</p><p> so l = 2.33 m</p><p> v 343 m s and f = = = 147 Hz l 2.33 m</p><p>For the string, l and v are different but f is the same.</p><p> l 0.400 m =d = 2NN 2</p><p> so l = 0.400 m T v=l f =(0.400 m) ( 147 Hz) = 58.8 m s = m 2 T=m v 2 = (9.00� 10- 3 kg m)( 58.8 m s) 31.1 N</p>