Question

A mass weighing 6 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numerically equal to β (β > 0) times the instantaneous velocity. Determine the values of the damping constant β so that the subsequent motion is overdamped, critically damped, and underdamped. (If an answer is an interval, use interval notation. Use g = 32 ft/s2 for the acceleration due to gravity.) (a) overdamped

Answer 1

Answer with Step-by-step explanation:

Let m be the mass attached and k be the spring constant and $\beta$ be the positive  damping constant.The Newton's second law for the system is given by

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

Where x(t) is the displacement from the equilibrium position.

The equation can be written as

$m\frac{d^2x}{dt^2}+\beta\frac{dx}{dt}+kx=0$

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0$

Weight=W=6 lb

s=2 feet

$g=32 ft/s^2$

$m=\frac{W}{g}=\frac{6}{32}=\frac{3}{16} slug$

$k=\frac{W}{s}=\frac{6}{2}=3 lb/ft$

$\frac{d^2x}{dt^2}+\frac{16\beta}{3}\frac{dx}{dt}+16 x=0$

$2\lambda=\frac{16\beta}{3}\implies \lambda=\frac{8\beta}{3}$

$\omega^2=\frac{k}{m}=16$

$\lambda^2-\omega^2=\frac{64\beta^2}{9}-16$

a.The motion is overdamped when $\lambda^2-\omega^2 >0$ or $\beta > \frac{3}{2}$

b.The motion is crirtically damped when  $\lambda^2-\omega^2 =0$ or $\beta = \frac{3}{2}$

c. The motion is underdamped when  $\lambda^2-\omega^2 <0$ or $\beta <\frac{3}{2}$

Answer 2

By using newton's second law for the system we got that for overdamped motion $\lambda^{2}-\omega^{2} > 0 \text { or } \beta > \frac{3}{2}$

What is  Newton's second law for the system ?

Newton's second law for the system is that acceleration of an object produced by a net force is directly proportional to magnitude of the net force, and inversely proportional to the mass of the object.

Given that

Weight$=W=6\ lb$

$s=2 \ ft$

$g = 32$ $ft/s^2$

We know that

$m \frac{d^{2} x}{d t^{2}}=-k x-\beta \frac{d x}{d t}$

Here x(t) is the displacement from the equilibrium

Now we can write it as

$\begin{aligned}&m \frac{d^{2} x}{d t^{2}}+\beta \frac{d x}{d t}+k x=0 \\&\frac{d^{2} x}{d t^{2}}+\frac{\beta}{m} \frac{d x}{d t}+\frac{k}{m} x=0\end{aligned}$

Now we can calculate m and k as

$\begin{aligned}&m=\frac{W}{g}=\frac{6}{32}=\frac{3}{16} \mathrm{slug} \\&k=\frac{W}{s}=\frac{6}{2}=3 \mathrm{lb} / \mathrm{ft}\end{aligned}$

Now we can write equations as

$\begin{aligned}&\frac{d^{2} x}{d t^{2}}+\frac{16 \beta}{3} \frac{d x}{d t}+16 x=0 \\&2 \lambda=\frac{16 \beta}{3} \Longrightarrow \lambda=\frac{8 \beta}{3} \\&\omega^{2}=\frac{k}{m}=16 \\&\lambda^{2}-\omega^{2}=\frac{64 \beta^{2}}{9}-16\end{aligned}$

Now for overdamped motion

$\lambda^{2}-\omega^{2} > 0 \text { or } \beta > \frac{3}{2}$

By using newton's second law for the system we got that for overdamped motion $\lambda^{2}-\omega^{2} > 0 \text { or } \beta > \frac{3}{2}$

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