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Unformatted text preview: SOLUTIONS – DECEMBER 5 15.7. IDENTIFY: Use Eq.(15.1) to calculate v . T = 1/ f and k is defined by Eq.(15.5). The general form of the wave function is given by Eq.(15.8), which is the equation for the transverse displacement. SET UP: v = 8.00 m/s, A = 0.0700 m, λ = 0.320 m EXECUTE: (a) v = f λ so f = v / λ = (8.00 m/s)/(0.320 m) = 25.0 Hz T = 1/ f = 1/ 25.0 Hz = 0.0400 s k = 2 π / λ = 2 π rad/0.320 m = 19.6 rad/m (b) For a wave traveling in the − xdirection, y ( x , t ) = A cos2 π ( x / λ + t / T ) (Eq.(15.8).) At x = 0, y (0, t ) = A cos2 π ( t / T ), so y = A at t = 0. This equation describes the wave specified in the problem. Substitute in numerical values: y ( x , t ) = (0.0700 m)cos(2 π ( x / 0.320 m + t / 0.0400 s)). Or, y ( x , t ) = (0.0700 m)cos((19.6 m − 1 ) x + (157 rad/s) t ). (c) From part (b), y = (0.0700 m)cos(2 π ( x / 0.320 m + t / 0.0400 s)). Plug in and x = 0.360 m t = 0.150 s: y = (0.0700 m)cos(2 π (0.360 m/0.320 m + 0.150 s/0.0400 s)) y = (0.0700 m)cos[2 π (4.875 rad)](4....
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This note was uploaded on 04/03/2008 for the course PHYSICS 140 taught by Professor Evrard during the Fall '07 term at University of Michigan.
 Fall '07
 Evrard

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