The displacement x(t) of a cart that is a part of a mass-spring system is descri

Question

The displacement x(t) of a cart that is a part of a mass-spring system is described by the differential equation. 12 d^2 x/dt^2 + 12 dx/dt + 3x = 0. with the following initial conditions: x (0) = 8, dx/dt (0) = -200. Find the maximum value of the VELOCITY of the cart (for positive values of time t), round if off to for significant decimal digits, and provide the result. A student found that the result was as follows _ (your numerical answer must be written here) Use, you must provide some intermediate results obtained by you while solving the problem above: 1) The characteristic (auxiliary) equation for the differential equation is as follows (mark a correct one): r^2 + 12 r + 12 = 0 12 r^2 + 12 r =0 12 r^2 + 12 r + 3 = 0 r^2 + 24r + 3 = 0 none of the above, specify:

Explanation / Answer

Given equation is 12x'' + 12x' + 3 = 0

The auxillary equation is 12r2 + 12r + 3 = 0.

=> 4r2 + 4r + 1 = 0

=> (2r+1)2 = 0

=> 2r = -1 => r = -1/2

Note that the root is repeated

Let the solution be x(t) = ae-t/2 + bte-t/2

When t = 0

x(0) = a = 8

Differentiating x(t)

=> x'(t) = (-1/2)ae-t/2 + be-t/2 + (-1/2)bte-t/2

When t = 0

x'(0) = -a/2 + b = -200

=> -8/2 + b = -200

=> -4 + b = -200

=> b = -200 + 4 = -196

The solution is x = 8e-t/2 -196te-t/2

Maximum velocity is when x''(t) = 0

Substituting in the given equation

=> 0 + 12x' + 3x = 0

=> 4x' + x = 0

=> 4((-1/2)ae-t/2 + be-t/2 + (-1/2)bte-t/2) + ae-t/2 + bte-t/2 = 0

=> -ae-t/2 + 4be-t/2 - bte-t/2 = 0

=> -a + 4b - bt = 0

=> -8 + 4*(-196) - (-196)t = 0

=> - 8 - 784 + 196t = 0

=> 196t = 792

=> t = 792/196 = 4.0408

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