A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 8 inches above the equilibrium position…

Question

asked Mar 14, 2021 146k views

1 Answer

Burger · Answer 1 · 2021-03-18T16:25:20+0000

Answer:

$x(t) = 8\cdot \cos 9.798\cdot t$

Explanation:

The equation of motion for the spring-mass system is:

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

Where:

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

- Spring constant, in
(lbm)/(s^(2)) .

- Mass, in
lbm .

The spring constant is:

$k = (m\cdot g)/(\Delta s)$

$k = ((24\,lbm)\cdot (32\,(ft)/(s^(2)) ))/((4)/(12) \,ft)$

$k = 2304\,(lbm)/(s^(2))$

The angular frequency is:

$\omega = \sqrt{(2304\,(lbm)/(s^(2)) )/(24\,lbm) }$

$\omega \approx 9.798\,(rad)/(s)$

The initial condition for the system is:

$x(0) = +8\,in$

$v(0) = 0\,(in)/(s)$

The function for speed is obtained by deriving the previous function:

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

The following expressions are formed by substituting all known variables:

$A \cdot \cos \phi = 8\,in$

$-(9.798\,(rad)/(s) )\cdot A \cdot \sin \phi = 0\,(in)/(s)$

The phase angle is found by dividing the initial velocity by the initial position:

$\tan \phi = 0$

$\phi = 0\,rad$

The amplitude is:

$A = 8\,in$

The equation of motion is:

$x(t) = 8\cdot \cos 9.798\cdot t$