Answer:Answer:
Initial value problem is:
u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0
u'(0) = 0.03m/s
Explanation:
The directions of Fd(t*) and U'(t*) are not specified in the question, so we'll take Fd(t*) to be negative and U'(t*) to be positive. This is due to the fact that the damping factor acts in the direction opposite the direction of the motion of the mass.
M = 5kg; L= 10cm or 0.1m;
F(t) = 10 sin(t/2) N ; Fd(t*) = - 2N
U'(t*) = 4cm/s or 0.04m/s
u(0) = 0
u'(0) = 3cm/s or 0.03m/s
Now, we know that W = KL.
Where K is the spring constant.
And L is the length of extension.
So, k = W/L
W= mg = 5 x 9.81 = 49.05N
So,k = 49.05/0.1 = 490.5kg/s^(2)
Now from spring damping, we know that; Fd(t*) = - γu'(t*)
Where,γ = damping coefficient
So, γ = - Fd(t*)/u'(t*)
So, γ = 2/0.04 = 50 Ns/m
Therefore, the initial value problem which describes the motion of the mass is;
5u'' + 50u' + 490u = (10 sin(t/2) N
Divide each term by 5 to give;
u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0
u'(0) = 0.03m/s
Explanation:
