Question

(1 point) A frictionless spring with a 6-kg mass can be held stretched 0.2 meters beyond its natural length by a force of 50 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 0.5 m/sec, find the position of the mass after tt seconds.

Answer 1

Answer:

x(t) = 0.077cos(6.455t)

Explanation:

If the spring can be stretched 0.2 m by a force of 50 N, then the spring constant is:

k = 50 / 0.2 = 250 N/m

The equation of simple harmonic motion is as the following:

$x(t) = Acos(\omega t - \phi)$

where $\omega = \sqrt{k/m} = \sqrt{250 / 6} = 6.455$

We also know that the initial velocity is 0.5 m/s, which is also the maximum speed at the equilibrium:

$v_{max} = A\omega$

$A = v_{max}/\omega = 0.5 / 6.455 = 0.077 m$

$\phi = 0$ is the initial phase

Therefore, the position of the mass after t seconds is

x(t) = 0.077cos(6.455t)