Question

A mass weighing 11 lb stretches a spring 4in. The mass is pulled down an additional 3 in and is then set in motion with an initial upward velocity of 5 ft/s. No damping is applied.
(a) Determine the position u of the mass at any time t. Use 32 ft/s² as the acceleration due to gravity. Pay close attention to the units.
(b) Determine the period, amplitude and phase of the motion.

Answer 1

Answer:

a) $x(t) = 2.845\cdot \cos \left(1.732\cdot t -0.473\pi \right)$, b) $T = 3.628\,s$, $A = 2.845\,ft$, $\phi = -0.473\pi$

Explanation:

a) The system mass-spring is well described by the following equation of equilibrium:

$\Sigma F = k\cdot x - m\cdot g = m\cdot a$

After some handling in physical and mathematical definition, the following non-homogeneous second-order linear differential equation:

$\frac{d^{2}x}{dt^{2}}+\frac{k}{m}\cdot x = g$

The solution of this equation is:

$x (t) = A\cdot \cos \left(\sqrt{\frac{k}{m} } \cdot t + \phi\right)$

The velocity function is:

$v(t) = \sqrt{\frac{k}{m} }\cdot A\cdot \sin \left(\sqrt{\frac{k}{m} }\cdot t +\phi \right)$

Initial conditions are:

$x(0\,s) = 0.25\,ft, v(0\,s) = -5\,\frac{ft}{s}$

Equations at $t = 0\,s$ are:

$0.25\,ft = A\cdot \cos \phi\\-5\,\frac{ft}{s} =\sqrt{\frac{k}{m} }\cdot A\cdot \sin \phi$

The spring constant is:

$k = \frac{11\,lbf}{0.333\,ft}$

$k = 33\,\frac{lbf}{ft}$

After some algebraic handling, amplitude and phase angle are found:

$\phi = -0.473\pi$

$A = 2.845\,ft$

The position can be described by this function:

$x(t) = 2.845\cdot \cos \left(1.732\cdot t -0.473\pi \right)$

b) The period of the motion is:

$T = \frac{2\pi}{\sqrt{\frac{k}{m} } }$

$T = 3.628\,s$

The amplitude is:

$A = 2.845\,ft$

The phase of the motion is:

$\phi = -0.473\pi$