Consider the equation for a non-linear oscillator where the force law contains b
ID: 1896939 • Letter: C
Question
Consider the equation for a non-linear oscillator where the force law contains both a linear and quadratic dependence on displacement from the equilibrium position. That is d2x/dt2 = -k/m x + b/m x2 The quantities k, m, and b completely characterise this problem. The ratio k/m has dimensions T-2 and b/m has dimensions L-1T-2. Thus a characteristic time scale for the problem is provided by the quantity The quantity Ohm = k/b is a characteristic length. So we may define a dimensionless time variable as Gamma = t/tau and a dimensionless length variable as X = x/Ohm. Using these definitions x becomes OhmX and d/dt = d/d Gamma d Gamma/dt = 1/tau d/d Gamma Making these substitutions in Equation(1) gives d2X/d Gamma 2 = -X(Gamma) + X2(Gamma) This is probably a good time to complete questions one and two below. What is special about the length k/b for Equation (1)? Show that Equation(1) becomes Equation(2) when you make the substitutions suggested in the text.Explanation / Answer
when x=k/b, the second derivative of x with respect to t becomes equal to 0.
Now,
dX/dT = (dX/dt)/(dT/dt)
Now, dX/dt = (1/)dx/dt
and dT/dt=(1/)
so, dX/dT = (/)dx/dt
So, d2X/dT2 = (d[(/)dx/dt]/dt)/(dT/dt)
=[(/)d2x/dt2]/(1/)
=(2/)d2x/dt2
=(2/)[-(k/m)x +(b/m)x2]
=(2/)[-(k/m)X + (b/m)(X)2]
Substituiting =(m/k) and =k/b,
=((m/k)/(k/b))[-(k2/mb)X + ((k2/mb)X2]
=(mb/k2)[-(k2/mb)X + ((k2/mb)X2]
=-X + X2 (reduced form)