Question

force of 400 N stretches a spring 2 m. A mass of 50 kg is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10 m/s. Find the equation of motion?

Answer 1

Answer:

x = -5 sin (2t)

Step-by-step explanation:

k is the spring stiffness.  The unstretched length of the spring is L.

When the mass is added, the spring stretches to an equilibrium position of L+s, where mg = ks.  When the mass is displaced a distance x (where x is positive if the displacement is down and negative if it's up), the spring is stretched a total distance s + x.

There are two forces on the mass: weight and force from the spring.  Sum of the forces in the downward direction:

∑F = ma

mg − k(s + x) = ma

mg − ks − kx = ma

Since mg = ks:

-kx = ma

Acceleration is second derivative of position, so:

-kx = m d²x/dt²

Let's find k:

F = kx

400 = 2k

k = 200

We know that m = 50.  Substituting:

-200x = 50 d²x/dt²

-4x = d²x/dt²

d²x/dt² + 4x = 0

This is a linear second order differential equation of the form:

x" + ω² x = 0

The solution to this is:

x = A cos (ωt) + B sin (ωt)

Here, ω² = 4, so ω = 2.

x = A cos (2t) + B sin (2t)

We're given initial conditions that x(0) = 0 and x'(0) = -10 (remember that down is positive and up is negative).

Finding x'(t):

x' = -2A sin (2t) + 2B cos (2t)

Plugging in the initial conditions:

0 = A

-10 = 2B

Therefore:

x = -5 sin (2t)