(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.

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:

where

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

is the initial phase

Therefore, the position of the mass after t seconds is

x(t) = 0.077cos(6.455t)

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.