string whose equilibrium position is x is displaced a short distance in the x di
ID: 2016216 • Letter: S
Question
string whose equilibrium position is x is displaced a short distance in the x direction to x + u ( x, t ) . Consider a short piece of siring of length I and use the definition ( 16.55 ) of Young's modulus YM to show that the tension is F = A YM u / x. where A is the cross sectional area of the string. [ If the string is already in tension in its equilibrium position, this F is the additional tension, that is. F = ( actual - equilibrium ) . ] ( b ) Now consider the forces on a short section of string dx and show that u obeys the wave equation with wave speed c = YM / q where Q is the density ( mass / volume ) of the string. stress = ( Young 's modulus ) times strain or df / A = Y M dl / l ( 16 .55 ) where YM is Young's modulus for the material of the wire.Explanation / Answer
a) consider a small part from x to x + dx, with length L = dx
at x, the short distance is u, at x + dx is u + du, so the moving distance for L is du = dL
dL/L = du/dx = ux
The force at x is F, where F/A = YM*dL/L = YM*ux, so F = AYM*ux
b) The net force on L is F (at x + dx) - F (at x)
= AYM*ux|x+dx - AYM*ux|x = AYM*ux/x * dx = AYM*2u/x2 * dx
the mass of L = dx is m = A*dx, the Newton's law is
m*2u/t2 = AYM*2u/x2 * dx
A*2u/t2 * dx = AYM*2u/x2 * dx
*2u/t2 - YM*2u/x2 = 0
2u/t2 - (YM/)*2u/x2 = 0
It is a wave equation with speed (YM/)
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