Answer to Question #249209 in Differential Equations for Micau

Question

A force of 2 pounds stretches a spring 1 foot. A mass weighing 3.2 pounds is attached to the
spring, and the system is then immersed in a medium that offers a damping force that is
numerically equal to 0.4 times the instantaneous velocity. (a) Find the equation of motion if the mass is initially released from rest from a point 1 foot above the equilibrium position. (b) Determine the position and velocity of the mass at t = sec? (c) Find the first time at which the mass passes through the equilibrium position heading upward.

Answer 1

We proceed to calculate and convert everything before proceeding

$\vec{F_{spring}}=-k\vec{x} \\ \implies k=\frac{F_{spring}}{x'}=\frac{mg}{x'} \\ k=\frac{2\,lb_f}{1\,ft}\times \frac{1\,ft}{0.3048\,m} \times \frac{4,4482\,N}{1\,lb_f} \\ \therefore k=29.1877\frac{N}{m} \\ \vec{F_{damping}}=-0.4\vec{v}=-[0.4\frac{lb_f}{s}]\frac{dx}{dt}=-b\frac{dx}{dt}$

$\sum F_{ext}=\vec{F_{damping}}+\vec{F_{spring}}=m\vec{a} \\ \implies m\cfrac{d^2x}{dx^2}=-b\cfrac{dx}{dt}-kx$

(a) Then, once we have the equation in SI units, we have the following expression:

$1.4515\cfrac{d^2x}{dx^2}+1.7793\cfrac{dx}{dt}+29.1877x=0 \\ \therefore \cfrac{d^2x}{dx^2}+1.2258\cfrac{dx}{dt}+20.1086x=0 \\ \therefore \cfrac{d^2x}{dx^2}+ \gamma\cfrac{dx}{dt}+ \omega^2_0 x=0$

We were able to find that the damping is light because

$\gamma^2/4=(1.2258^2)/4<\omega^2_0$

Now we proceed to find the solution for x(t) and from that expression we can find the velocity at any time:

$x_{(t)} = A_0 \exp(−γ t/2) \cos {(ωt)} \\ v_{(t)} = x'_{(t)}= -A_0 \exp(−γ t/2)[ ω \sin {(ωt)} +(γ /2)\cos {(ωt)} ] \\ A_0=\text{2 ft = 0.6096 m} \\ ω = (ω^2_o− γ^2/4)^{1/2}= (20.1086^2−1.2258^2/4)^{1/2}=20.0992\,s^{-1}$

(b) After substituion, we find:

$x_{(t)} = (0.6096\,m) \exp(− 10.0496t) \cos {(20.0992t)} \\ v_{(t)} =-(0.6096\,m) \exp(−10.0496t)[ (20.0992\,s^{-1}) \sin {(20.0992t)} +(10.0496\,s^{-1})\cos {(20.0992t)} ]$

(c) The mass goes up when $\cos(\omega t)=\cos(3\pi /2) \implies t=\cfrac{3\pi}{2\omega }$

We use the argument 3π/2 for the cosine function because that is the one that defines the displacement when the mass is going up (on the other hand when ωt=π/2 the mass is going down). Then, the time we were looking for will be:

$t=\cfrac{3\pi}{2(20.0992\,s^{-1})}=0.2344\,s$

Reference:

• King, G. C. (2013). Vibrations and waves. John Wiley & Sons.