Answer to Question #271802 in Differential Equations for Pathuri Nitish

Question

A 16 lb weight is suspended from a spring having spring constant 5 lbft. Assume that an external force given by 24 sin (10t) and a damping force with damping constant 4, are acting on the spring Initially the weight is at rest at its equilibrium position. Find the position of the weight at any time. Find the steady state solution. Find the amplitude, period and frequency of the steady state solution. Determine the velocity of the weight at any time

Answer 1

Solution;

Given;

$w=16lb$

$F(t)=24sin(10t)$ $k=5lbft$

$c=4$

The differential equation;

$\frac{w}{g}\ddot{x}=-c\dot{x}-kx+F(t)$

By direct substitution;

$\frac{32}{16}\ddot{x}+4\dot{x}+5x=24sin(10t)$

Rewrite as;

$\frac{d^2x}{dt^2}+8\frac{dx}{dt}+10x=48sin10t$

Which is a second order Linear differential equation;

The auxiliary equation is;

$r^2+8r+10=0$

The roots;

$r=\frac{-8_-^+\sqrt{64-40}}{2}$

$r_1=-4-\sqrt{6}$

$r_2=-4+\sqrt{6}$

The complementary solution is;

$x_c(t)=c_1e^{r_1t}+c_2e^{r_2t}$

The complementary solution is;

$x_p(t)=\frac{1}{D^2+8D+10}48sin10t$

$x_p(t)=\frac{48}{-100+8D+10}sin10t=\frac{48(8D+90)}{(8D+90)(8D-90)}sin10t=\frac{48(8D+90)}{64D^2-8100}sin10t$

Therefore we can write;

$x_p(t)=\frac{48(80cos10t+90sin10t)}{64(-100)-8100}$

$x_p(t)=\frac{-48}{1450}(8cos10t+9sin10t)$

The position of the weight at any time t;

$x(t)=x_c(t)+x_p(t)$

$x(t)=c_1e^{r_1t}+c_2e^{r_2t}-\frac{48}{1450}(8cos10t+9sin10t$

Using the initial conditions;

$x=0,t=0$

We have;

$0=c_1+c_2-\frac{48}{1450}×8$

$c_1+c_2=\frac{384}{1450}$ ...(a)

Differentiating x(t) w.r.t 't' we get;

$\dot{x}(t)=r_1c_1e^{r_1t}+r_2c_2e^{r_2t}-\frac{48}{1450}(-8sin10t+9cos10t)$

Applying the initial conditions;

$\dot{x}=0,t=0$

$0=r_1c_1+r_2c_2-\frac{432}{1450}$ ...(b)

Solving (a) and (b) gives;

$x(t)=0.96e^{-1.55t}-0.69e^{-6.45t}-0.298sin10t-0.265cos10t$

The transient solution is;

$0.96e^{-1.55}-0.695e^{-6.45t}$

The steady state solution is;

$-0.298sin10t-0.265cos10t$

$=0.397sin(10t+3.87)$

From the steady state part;

$Amplitude=0.397ft$

$Period=\frac{2π}{w}=\frac{2π}{10}=0.6283sec$

$Frequency=\frac{1}{T}=1.591$

Velocity of the weight at any time (t);

$\dot{x}(t)=-1.5e^{1.55t}+4.48e^{-6.45t}-2.98cos10t+2.65sin10t$