Question

A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 12 m/s. Find the equation of motion. x(t)

Answer 1

Answer:

$x(t) = (6\,m)\cdot \cos \left[\left(2\,\frac{rad}{s} \right)\cdot t-0.5\pi \right]$

Explanation:

This problem describes a system experimenting a simple harmonic motion. The spring constant is determined by the Hooke's Law:

$k = \frac{F}{\Delta x}$

$k = \frac{880\,N}{4\,m}$

$k = 220\,\frac{N}{m}$

The angular frequency is:

$\omega = \sqrt{\frac{k}{m} }$

$\omega = \sqrt{\frac{220\,\frac{N}{m} }{55\,kg} }$

$\omega = 2\,\frac{rad}{s}$

The position and velocity functions for the motion of the mass-spring system is:

$x(t) = A\cdot \cos(\omega\cdot t +\phi)$

$v(t) = -\omega \cdot A\cdot \sin (\omega\cdot t + \phi)$

The initial condition for the system are:

$x(0) = 0$, $v(0) = 12\,\frac{m}{s}$

Equation at initial time are:

$0 = A\cdot \cos \phi$

$12 = -2 \cdot A \cdot \sin \phi$

By dividing the first equation by the second one:

$\cot \phi = 0$

Which corresponds to $\phi = -0.5\pi$

The amplitude is:

$A = -\frac{6}{\sin (-0.5\pi)}$

$A = 6\,m$

The equation of motion is:

$x(t) = (6\,m)\cdot \cos \left[\left(2\,\frac{rad}{s} \right)\cdot t-0.5\pi \right]$