Question

A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. If the mass is set in motion in a medium that imparts a damping force numerically equal to 16 times the velocity, the correct differential equation for the position x (t ), of the mass at a function of time, t is

Answer 1

Answer:

d²x/dt² = - 4dx/dt - 4x is the required differential equation.

Explanation:

Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,

F = mg

kx = mg

k = mg/x

= 4 kg × 9.8 m/s²/2.45 m

= 39.2 kgm/s²/2.45 m

= 16 N/m

Now the drag force f = 16v where v is the velocity of the mass.

We now write an equation of motion for the forces on the mass. So,

F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass

-kx - 16v = 4a

-16x - 16v = 4a

16x + 16v = -4a

4x + 4v = -a where v = dx/dt and a = d²x/dt²

4x + 4dx/dt = -d²x/dt²

d²x/dt² = - 4dx/dt - 4x which is the required differential equation