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Reflection and Transmission What force must be applied at end of a string to launch a pulse? F THERE ARE TWO FORCES TO CONSIDER !!!!!! Force on a string segment Dx: (from when we derived the wave equation) As Dx goes to zero, Fy and mass go to zero, but acceleration of the string segment remains finite. It had better – we need a force to move the string to get a wave equation ! T Vertical force at the string end: Fy T sin (Tension force at end) The Impedance Creed The string end is an interface. It has no width. It has no mass. There is no limit to apply. It cannot have a net force applied to it. A net force would be equivalent to an infinite stress. A net force at the end would rip the string. The driver must therefore balance this force. If y = f(x-vt): Force needed is proportional to the transverse velocity, like a damping force! Constant of proportionality is the characteristic impedance: Z T The change in impedance at an interface determines the amounts of reflection and transmission. Z1 T1 f1 x vt g1 x vt Z 2 T 2 f 2 x vt The string must be continuous at the boundary: f1 t g1 t f 2 t The force must balance at the boundary: f1 g1 Fy T T x x f 2 Fy T x f g Fy Z1 1 1 t t f Fy Z 2 2 t f1 g1 f 2 Z1 Z2 t t t Integrate with respect to time: Z1 f1 t Z1 g1 t Z 2 f 2 t Z1 f1 t Z1 g1 t Z 2 f1 t Z 2 g1 t Z1 g1 t Z 2 g1 t Z1 f1 t Z 2 f1 t r g1 t Z1 Z 2 f1 t Z1 Z 2 reflection coefficient Z1 f1 t Z1 f 2 t Z1 f1 t Z 2 f 2 t Z1 f 2 t Z 2 f 2 t 2 Z1 f1 t f 2 t 2 Z1 t f1 t Z1 Z 2 transmission coefficient Into more dense medium Into less dense medium 1 2 1 2 Impedance Matching: No Reflection! Z1 Z 2 T11 T11 (not really an interface) T11 T2 2