A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10 sin (t/2) N (newtons) and moves in a medium that imparts a viscous force of 2 N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass.

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