Suppose a mass ( m ) is hung on a spring with spring constant ( k ). If the spring is compressed or stretched and then released, the mass will oscillate up and down. Because of friction, the oscillation will be damped: eventually the motion will cease. The damping will depend on the amount of friction; for example, if the system is suspended in oil the motion will cease sooner than if the system is in air. Using some simple physics, it is not hard to see that the position of the mass is described by this differential equation: [ m ddot{y}+b dot{y}+k y=0 ] Using ( m=1 ) and ( k=5 ) compare the motion of the mass in oil and air. Given that the damping coefficient of air ( b=1 ) and damping coefficient of oil ( b=3 ). Suppose we know that ( y(0)=1 ) and ( dot{y}(0)=2 ). Use system of difference recurrence equations to plot response for of this homogeneous equation for 10 seconds with the time interval of ( 1 mathrm{msec} ). How many secons sooner will the motion cease in oil compared to the system is in air? Submit, 1) answer, 2) Matlab code, 3) graph(s)
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Step 1: Find the solution for the air case: Given the second order differential equation \(y'' + y' + 5y = 0\) with initial data, we can find the general solution to be \(y(t) = e^{-\frac{1}{2}t}(A\cos(\frac{\sqrt{19}}{2}t) + B\sin(\frac{\sqrt{19}}{2}t))\). Show more…
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