Answer to Question #181855 in Differential Equations for Miriam Vogel

Question

A particle is attached to the lower end of a spring, the upper end of which oscillates about a point O. The motion of the particle can be modelled by the equation

d^2x/dt^2 + 25x = 0.5sint.

where x is the displacement of the particle from its equilibrium point. When t = 0, x = 0 and the particle is at rest.

(a) Solve this differential equation to find x in terms of t and describe briefly the motion of the particle.

In order to damp the oscillations the particle is submerged in liquid and the motion of the particle can be modelled as

d^2x/dt^2 + kdx/dt + 25x = 0.5sint, where k is a constant.

(b) Explain why k must be positive. Give the range of values of k for which the system will be underdamped.

Answer 1

Solution with correction

(a) For homogeneous equation d²x/dt² + 25x = 0 let’s solve characteristic equation r²+25 = 0.

r = ±5i => general solution of homogeneous equation is x₀(t) = Asin(5t)+Bcos(5t), whre A,B are arbitrary constants.

Partial solution of nonhomogeneous equation d²x/dt² + 25x = 0.5sint is x₁(t) = Csint. Substitution into equation gives -C+25 = 0.5 => C = 24.5.

So general solution of equation d²x/dt² + 25x = 0.5sint is

x(t) = x₀(t)+x₁(t) = Asin(5t)+Bcos(5t)+24.5sint

If for t = 0, x = 0 and the particle is at rest, x(0) = 0 and x’(0) = 0. => B = 0, 5A+24.5 = 0 => A = -4.9, B = 0 => x(t) = -4.9sin(5t)+24.5sint

The motion of the particle is superposition of two vibratios with periods 2π and 2π/5 and amplitudes 24.5 and 4.9

(b) For homogeneous equation d²x/dt² + kdx/dt + 25x = 0 characteristic equation is r²+kr+25 = 0.

$r_{1,2} = -\frac{k}{2}\pm\sqrt{(\frac{k}{2})^2-25}$

Mass for underdamped system oscillates within a decaying envelope. So r_1,2 are to be complex numbers with negative real part. Therefore k>0 and 25- (k/2)² >0. From second inequality |k|<10. So system will be underdamped for 0<k<10.