Case I: The force acting on an object of mass m is $F(x) = F_o(1-e^{\alpha x})$

Case II: The force acting on an object of mass m is $F(x) = F_o(1-e^{-\alpha x})$

where $F_o$ and $\alpha$ are positive constants. In which case can the object oscillate near the equilibrium and what is the oscillation frequency?


The total force on the object is: $$ F(x, x', t) = -kx + F_o(1-e^{\alpha x})$$ and can be rewritten as: $$ x'' + w^2 x = F_o(1-e^{\alpha x})$$ where $w^2 = -k/m$

This is an inhomogeneous linear differential equation. I first solve for the complementary solution, $x_c$, for $x'' + w^2x = 0$. This gives me: $$x_c = Ae^{0x} + Be^{-w^2x} = A + Be^{-w^2x}$$

I can then determine the particular solution, $x_p$ for: $$ 1)x'' + w^2 x = F_o$$ $$ x_{p1} = \frac{F_o}{w^2} $$ $$2) x'' + w^2 x = -F_oe^{\alpha x}$$ $$ x_{p2} = \frac{-F_oe^{\alpha x}}{\alpha ^2 + 1}$$

For a general solution of: $$ x = A + Be^{-w^2 x} + \frac{F_o}{w^2} + \frac{-F_oe^{\alpha x}}{\alpha ^2 + 1}$$

This raises two questions:

1 - Is this solution correct?

2- Once I solve for $F(x) = F_o(1-e^{-\alpha x})$, how do I determine the case in which the object oscillates near the equilibrium? I suppose I would need to minimize the amplitude? I'm not sure how to approach this.


I don't think is the correct way to go about the problem. In your first case you've just arbitrarily added another force, which is not what the particle is asking.

The way these problems are usually done is you should do perturb the correct path of the particle x by dx so the perturbed path is x+dx where dx << x. Plug this in the equation then do some kind of binomial expansion to order dx and simple harmonic motion should pop out of the correct case.

  • $\begingroup$ I'm afraid I don't really understand what you're trying to say. How will adding velocity (dx) to the position help me find harmonic motion? $\endgroup$ – Alex Jan 28 '15 at 18:03

An object that oscillates near equilibrium will essentially follow the harmonic motion F = -kx, and at equilibrium x = 0. I take the derivatives for a) and b) and solve for frequency, where $ w = \sqrt{k/m}$:

a) $ (F_o - F_o e^{\alpha x})\frac{d}{dx} = {-kx}\frac{d}{dx} $

$ -F_o \alpha e^{\alpha x} = -k $

$k = F_o \alpha e^{\alpha x}$, and for x = 0 we have: $$ w = \sqrt{\frac{F_o \alpha}{m}} $$

b) $ (F_o - F_o e^{-\alpha x})\frac{d}{dx} = {-kx}\frac{d}{dx} $

$ F_o \alpha e^{\alpha x} = -k $

$k = -F_o \alpha e^{\alpha x}$, and for x = 0 we have: $$ w = \sqrt{\frac{-F_o \alpha}{m}} $$ which results in an imaginary number, so it doesn't work given the question at hand. Therefore, the object can only oscillate near the equilibrium position and a), and the frequency w is given as $$ w = \sqrt{\frac{F_o \alpha}{m}} $$


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