A body of mass m kg attached to a spring moves with friction. The motion is desc

Question

A body of mass m kg attached to a spring moves with friction. The motion is described by the second Newton's law: m(d^2y/dt^2) + a(dy/dt) + ky = 0 where y is the body displacement in m, t is the time in s, a > 0 is the friction coeffcient in kg/s and k is the spring constant in kg/s2. Assuming m = 1 kg and k = 4 kg/s2, find: a) What is the range of values of a for which the body moves (i) with oscillations, (ii) without oscillations? b) Find the general solution for any a < 4. (Your solution should be a formula depending on the parameter a.) Proof that it follows from the solution obtained that the body slows down to a virtually rest state at large time (i.e. when t --> infinity)? c) Find the particular solution for a = 4 subject to the initial conditions y(0) = 0, dy/dt = 1 m/s at t = 0. Plot this solution and determine the largest displacement of the mass using calculus.

Explanation / Answer

a).

m*(d2y/dt2) + a*(dy/dt) + ky = 0

or my'' + ay' + ky = 0

or y'' + (a/m)*y' + (k/m)y = 0

m= 1 kg ; k =4 kg/s2

therefore, y'' + a*y' + 4y = 0

solving this differential equation from the D-operator method:

characteristics equation,

D2 + a*D + 4 = 0

roots of the equation, D1 ,D2 = [-a ±(a2-4*4)]/2 = -a/2 ±((a/2)2 - 4)

general solution of such equation:

roots = ±i

then, solution = et*(Acost + Bsint)

i). with oscillations:

if < 0 , then oscillation takes place but this is damping oscillation

= 0 , no oscillation will takes place

> 0 , oscillation will takes place but this is over-damping oscillation.

here given that motion is resisted by friction force, therefore over-damping is not possible.

hence,

for oscillation, ((a/2)2 - 4) < 0

or a2-16<0

or (a+4)(a-4) < 0

then, solution is -4<a<4

but a>0 is given,

therefore range for a with oscillation:   0<a<4

ii). without oscillation:

((a/2)2 - 4) = 0

a>0, therefore,

a = 4

b). general solution for a<4 :

roots = -a/2 ±((a/2)2 - 4)

or -a/2 ±i

where = (1/2)*(a2-16)

hence general solution, y(t) = e-at/2*(Acost + Bsint)

y(infinity) = 0

at time tending to infinity, displacement will be 0.

c).

a=4

then = 0

therefore,

D2 + a*D + 4 = 0 or D2 + 4D + 4 = 0

that means, above equation has repeated roots and its value = -2

solution of the differential equation is:

y(t) = (A+t*B)e-2t

y'(t) = (-2A+B-2Bt)e-2t

intial condition,

y(t=0) = 0 ; y'(t=0) = 1

therefore,

y(t) = (A+t*B)e-2t

0 = A+0

A = 0

y'(t) = (-2A+B-2Bt)e-2t

1 =-0+B -0

B=1

therefore solution, y(t) = te-2t

largest displacement of the mass:

y'(t) = (1-2t)e-2t

for maximum, dy/dt = 0

y'(t) = 0 =(1-2t)e-2t

after solving this, we get;

t = 0.5 second

and maximum displacement = 0.5*e-1 = 1/(2e) m

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