s18 - P18.4 Suppose the waves are sinusoidal. 00 n 00 n 0...

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P18.4 Suppose the waves are sinusoidal. The sum is () ( ) ( ) 4.00 cm si n4 . 00 cm si n9 0 0 kx t t ωω . +− + ° 24.00 cm si 5 . 0c o s 4 5 .0 t ω ° So the amplitude is 8.00 cm cos45. 05 . 66 cm °= . P18.6 00 2c o s 2 AA φ ⎛⎞ = ⎜⎟ ⎝⎠ so 1 1 cos 60.0 2 23 π = = Thus, the phase difference is 2 120 3 = This phase difference results if the time delay is 1 3 33 T f v λ == Time delay 3.00 m 0.500 s 32.00 m s P18.9 We suppose the man’s ears are at the same level as the lower speaker. Sound from the upper speaker is delayed by traveling the extra distance 22 L dL + . He hears a minimum when this is ( ) 2 1 2 n with 1,2,3, n = K Then, ( ) 12 nv LdL f +−= ( ) 2 2 2 2 2 2 2 21 2 Ld L f n L f f dn v f Ln f += + −− ++ K v L This will give us the answer to (b). The path difference starts from nearly zero when the man is very far away and increases to d when 0 L = . The number of minima he hears is the greatest integer solution to ( ) d f n = greatest integer 1 2 df v + . continued on next page 523
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524 Superposition and Standing Waves (a) ( ) ( ) 4.00 m2 0 0 s 11 2.92 2 330 m s 2 df v += He hears two minima. (b) With 1 n = , () 2 22 2 2 1 24 . 00 m3 3 0 m s 4 2 0 0 s 2 1 2 330 m s 200 s 9.28 m dv f L vf L −− == = with 2 n = ( ) 2 2 32 1.99 m 232 f L . P18.16 0 2s i nc o s yA k x t ω = 2 2 0 2 i o s y Ak k x t x =− 2 2 0 2 2 si o s y A kx t t Substitution into the wave equation gives 00 2 1 i o s 2 s i o s A k k xt A k v ωω ⎛⎞ −= ⎜⎟ ⎝⎠ This is satisfied, provided that v k = P18.18 (a) The resultant wave is i o s 2 2 k x t φφ =+ The nodes are located at 2 k xn φ π so 2 n x k k πφ which means that each node is shifted 2 k to the left. (b) The separation of nodes is 1 2 2 n k kk k ⎤⎡ Δ= + ⎥⎢ ⎦⎣ 2 x k λ Δ= = The nodes are still separated by half a wavelength.
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s18 - P18.4 Suppose the waves are sinusoidal. 00 n 00 n 0...

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