Question

A 0.200-kg mass is attached to the end of a spring with a spring constant of 11 N/m. The mass is first examined (t = 0) when the mass is 17.0 cm from equilibrium in the positive x-direction, and is traveling at 2.0 m/s in the positive x-direction.
a) Write an equation x(t) that describes the position of this mass as a function of time. Express this function in terms of numerical values, trigonometric functions and the time variable "t".b) Repeat for v(t), the speed of the mass as a function of time.c) Repeat for a(t), the acceleration of the mass as a function of time.

Answer 1

Answer:

a) x (t) = 0.3187 cos (7.416 t + 1.008) ,  b)  v = -2,363 sin (7,416 t + 1,008)

c)  a = - 17.52 cos (7.416t + 1.008)

Explanation:

The spring mass system creates a harmonic oscillator that is described by the equation

    x = Acos (wt + φ)

Where is the amplitude, w the angular velocity and fi the phase

a) Let's reduce the SI system

    x = 17.0 cm (1 m / 100 cm) = 0.170 m

The angular velocity is given by

      w = √ (k / m)

      w = √ 11 / 0.200

      w = 7.416 rad / s

Let's look for the terms of the equation with the data for time zero (t = 0 s)

      0.170 = A cos  φ

Body speed can be obtained by derivatives

      v = dx / dt

      v = -A w sin (wt + φ)

     2.0 = -A 7.416 sin φ

Let's write the two equations

     0.170 = A cos φ

     2.0 / 7.416 = -A sin φ

Let's divide those equations

    tan φ= 2.0 / (7.416 0.170)

     φ= tan⁻¹ (1,586)

     φ= 1.008 rad

We calculate A

   A = 0.170 / cos φ

   A = 0.170 / cos 1.008

   A = 0.3187 m

With these values ​​we write the equation of motion

    x (t) = 0.3187 cos (7.416 t + 1.008)

b) the speed can be found by derivatives

      v = dx / dt

      v = - 0.3187 7.416 sin (7.416 t +1.008)

      v = -2,363 sin (7,416 t + 1,008)

c) the acceleration we look for conserved

    a = dv / dt

    a = -2,363 7,416 cos (7,416 t + 1,008)

    a = - 17.52 cos (7.416t + 1.008)