Question

A block weighing 8 pounds is attached to a spring. When set in motion, the spring/massvsystem exhibits simple harmonic motion. (a) Determine the equation of motion if the spring constant is 1 lb/ft and the mass is initially released from a point 6 inches below the equilibrium position with a downward velocity of 3/2 ft/sec. Express the equation of motion in alternative form. (b) What are the amplitude and period of motion? (c) What is the position and velocity of the mass at t = pi/ 4 sec?

Answer 1

a)

$m\frac{d^2x}{dt^2}=-kx$

$8\frac{d^2x}{dt^2}+x=0$

$8k^2+1=0$

$k=\pm i/2\sqrt{2}$

$x(t)=c_1cos(t/(2\sqrt{2}))+c_2sin(t/(2\sqrt{2}))$

From the initial conditions:

$x(0)=6$ in$=1/2$ ft

$c_1=1/2$

$x'(0)=3/2$ ft/s

$x'(t)=-\frac{1}{4\sqrt{2}}sin(\frac{t}{2\sqrt{2}})+\frac{c_2}{4\sqrt{2}}cos(\frac{t}{2\sqrt{2}})$

$c_2=\frac{3}{2}\cdot 4\sqrt{2}=6\sqrt{2}$

we can find equation of motion:

$x(t)=\frac{1}{2}cos(t/(2\sqrt{2}))+6\sqrt{2}sin(t/(2\sqrt{2}))$

b)

Amplitude:

$A=\sqrt{c_1^2+c_2^2}=\sqrt{(1/2)^2+2\cdot6^2}=\sqrt{289}/2=17/2$

The phase angle:

$tan\phi=c_1/c_2=1/(12\sqrt{2})$

$\phi=tan^{-1}(1/(12\sqrt{2}))=0.588$

the equation of motion in alternative form:

$x(t)=\frac{17}{2}sin(\frac{t}{2\sqrt{2}}+0.588)$

period of motion:

$T=2\pi/\omega=2\pi\cdot2\sqrt{2}=4\sqrt{2}\pi$

c)

the position:

$x(\pi/4)=\frac{1}{2}cos(\pi/(8\sqrt{2}))+6\sqrt{2}sin(\pi/(8\sqrt{2}))=0.481+2.326=2.807$ ft

the velocity:

$v(\pi/4)=-\frac{1}{4\sqrt{2}}sin(\frac{\pi}{8\sqrt{2}})+\frac{3}{2}cos(\frac{\pi}{8\sqrt{2}})=-0.048+1.020=0.972$ ft/s