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

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)

Step-by-step 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)