A 50g mass stretches a spring 10cm. As the system moves through the air, a resis

Question

A 50g mass stretches a spring 10cm. As the system moves through the air,
a resistive force is supplied that is proportional to, but opposite the velocity,
with magnitude .01v. The system is hooked to a machine that applies a driving
force to the mass that is equal to F(t) = 5 cos(4:4t). If the system is started
from equilibrium (no displacement, no velocity) SET UP BUT DO NO SOLVE
an ODE and two ICs to describe the motion . (Remember 1000g=1kg and
100cm=1m).
Find the position of the mass as a function of time, i.e.
solve the ODE

Explanation / Answer

0.05*9.8 = k (0.1); gives k = 4.9

The ODE is as follows:

0.05(d2x/dx^2) + 0.01(dx/dt) +4.9 x = 5 cos(4t) ; the question gives 5 cos(4:4t), hope thats is a typing mistake.

on solving using laplace we get the equation;

X(s) *(0.05s^2 +0.01s+ 4.9) = (5*s)/(s^2 + 16)

implies X(s) = [(5*s)/(s^2 + 16)] /(0.05s^2 +0.01s+ 4.9)

solving ; ie taking the inverse laplace we get ;

x(t) = (51250*cos(4*t))/42029 + (500*sin(4*t))/42029 - (51250*exp(-t/10)*(cos((9799^(1/2)*t)/10) + (57*9799^(1/2)*sin((9799^(1/2)*t)/10))/401759))/42029

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