Question

Differential Equations - Solve "A 0.25 kg mass is horizontally attached to a spring with a stiffness 4 N/m. The damping
constant b for the system is 1 N-sec/m. If the mass is displaced 0.5 m to the left and given an
initial velocity of 1 m/sec to the left, find the equation of motion. What is the maximum
displacement that the mass will attain?"

Answer 1

The D.E. would be mx'' + bx' + kx = 0, where m is the mass, b is the damping and k is the stiffness. So we have

0.25x'' + 1x' + 4x = 0
x'' + 4x' + 16x = 0
The auxiliary equation is u^2 + 4u + 16 = 0. Solving it we get u = -2 +- 2sqrt(3)i. Hence we get that

x(t) = e^(-2t) (A cos (2sqrt(3)x) + B sin (2sqrt(3)x))

Now we have that, initially, the mass is displaced by 0.5m to the left. Assuming that the left is the positive direction, we have x(0)=0.5. Hence

x(0) = 0.5 = e^0 (A cos 0 + B sin 0). Hence A = 0.5.

Also, the initial velocity is 1m/s to the left; hence x'(0) = 1.

x'(t) = e^(-2t) (2sqrt(3) (-A sin (2sqrt(3)x) + B cos (2sqrt(3)x)) - 2e^(-2t) (A cos (2sqrt(3)x) + B sin (2sqrt(3)x)) (using the product rule)

x'(0) = 1 = e^0 (2sqrt(3) (-A sin 0 + B cos 0) - 2e^0 (A cos 0 + B sin 0)

1 = 2sqrt(3) B - 2A
1 = 2sqrt(3) B - 1
2 = 2sqrt(3) B. Hence B = 1/sqrt(3).

Therefore the particular solution is:

x(t) = e^(-2t) (0.5 cos (2sqrt(3)x) + (1/sqrt(3)) sin (2sqrt(3)x))

The maximum displacement occurs when x'(t) = 0, i.e. when
e^(-2t) (2sqrt(3) (-A sin (2sqrt(3)x) + B cos (2sqrt(3)x)) - 2e^(-2t) (A cos (2sqrt(3)x) + B sin (2sqrt(3)x)) = 0

The exponentials can be extracted as common factors, and they can never be zero, so we end up with

(-sqrt3 sin (2sqrt(3)x) + 2 cos (2sqrt(3)x) - cos (2sqrt(3)x) + 2/sqrt(3) sin (2sqrt(3)x)) = 0
cos(2sqrt(3)x) - (1/sqrt3) sin(2sqrt(3)x) = 0
cos(2sqrt(3)x) = (1/sqrt3) sin(2sqrt(3)x)
sqrt3 = tan(2sqrt(3)x)

Hence 2sqrt(3)x = pi/3 + n pi, n is any natural number
x = pi / (6sqrt(3)) + n pi / (2sqrt(3)), n is any natural number.

Answer 2

Sorry, t = pi / (6sqrt3) not x.

The maximum displacement would be x(t) for this value of t, i.e.

x = e^(-2pi / (6sqrt3)) (0.5 cos (pi/3) + (1/sqrt(3)) sin (pi/3))
x = e^(-2pi / (6sqrt3)) (0.5 x 0.5 + (1/sqrt3)(\sqrt3 / 2))
x = 0.75 e^(-2pi / (6sqrt3)) is the largest displacement.

Answer 3

The value of A is 0.5 or - 0.5.....