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

`x(t) = (6\,m)\cdot \cos \left[\left(2\,(rad)/(s) \right)\cdot t-0.5\pi \right]`

Step-by-step explanation:

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

`k = (F)/(\Delta x)`

`k = (880\,N)/(4\,m)`

`k = 220\,(N)/(m)`

The angular frequency is:

`\omega = \sqrt{(k)/(m) }`

`\omega = \sqrt{(220\,(N)/(m) )/(55\,kg) }`

`\omega = 2\,(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\,(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 = -(6)/(\sin (-0.5\pi))`

`A = 6\,m`

The equation of motion is:

`x(t) = (6\,m)\cdot \cos \left[\left(2\,(rad)/(s) \right)\cdot t-0.5\pi \right]`