(1 point) A spring with a 5-kg mass and a damping constant 4 can be held stretch

Question

(1 point) A spring with a 5-kg mass and a damping constant 4 can be held stretched 2.5 meters beyond its natural length by a force of 10 newtons. Suppose the spring is stretched 5 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c -4mk? general form cie cos(t)esin(ot) a= m2kg? /sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t with the 6- Note: You can eam partial credit on this problem,

Explanation / Answer

Solution:

m*d2x/dt2 + Kv*dx/dt + K*x=0

where Kv is viscous damping constant and K is spring constant.

Laplace Transform:

m[s2*X(s) - sx(0) - x'(0)] + Kv[s*X(s) - x(0)...

X(s){m*s2 + Kv*s + K} = m[s*(0) + x'(0)] + Kx*(...

X(s) = {m[s*(0) + x'(0)] + Kx*(0)}/{m*s2+Kv...

X(s) = (A*s+B)/{m*s2 + Kv*s + K}

Partial fractions can reveal values for A and B and then the time domain solution can be derived.

However for critical damping it is sufficient to examine the denominator.

If the 'discriminator' equals zero then there is a single root and this corresponds to 'critical damping'.

So, as for normal quadratic equation 'b2 - 4ac = 0' is required.

This is: Kv2 - 4*m*K = 0

i.e. m = Kv2/(4*K)

Kv = 5 N/m/s (presumably, given in question); K=10/2.5 = 4 N/m

m = 52/(4*5) = 25/20 = 1.25 Kg

I believe the '5 meter' release information is redundant.

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