Answer to Question #163888 in Differential Equations for Janet

Question

A particle of mass 5kg moves along x-axis under the influence of two forces;force of attraction to the origin which is numerically equal to 40 times the instantaneous distance from the origin and damping force is proportional to instantaneous speed; when the speed is 10m/s the damping is 200. Assuming that the particle from rest at a distance of 20cm from the origin then

a)Write the differential equations

b)Find the position at any given time t

c) Determine the amplitude, frequency and period of the damping oscillation.

Answer 1

Solution:

(a):

Given, when the speed is 10m/s the damping is 200 N.

Constant of proportionality $\dfrac{200}{10}=20$ N-s/m

So, making an equation using given forces as follows:

$5a=-40x-20v$ , where a is acceleration, x is displacement and v is velocity.

$\Rightarrow a+4v+8x=0 \\ \Rightarrow \dfrac{d^2x}{dt^2}+4\dfrac{dx}{dt}+8x=0$

Required differential equation is $\dfrac{d^2x}{dt^2}+4\dfrac{dx}{dt}+8x=0$

(b):

Now, solving this differential equation to get its position.

Its auxiliary equation is $D^2+4D+8=0$

$\Rightarrow D=\dfrac{-4\pm\sqrt{4^2-4(8)}}{2}=\dfrac{-4\pm\sqrt{-32}}{2}=\dfrac{-4\pm4i\sqrt{2}}{2}$

$\Rightarrow D=-2\pm2\sqrt{2}i$

Thus, its general solution is: $x(t)=e^{-2t}(c_1\cos 2\sqrt2t+c_2\sin 2\sqrt2t)$

Or $x(t)=Re^{-2t}\cos (2\sqrt2t-\theta)$ ...(i)

Now, put $t=0,x=20\ cm=0.2\ m$

$0.2=Re^{0}\cos (0-\theta) \\ \Rightarrow R\cos \theta=0.2$ ...(ii)

Differentiating (i) w.r.t t, we get,

$x'(t)=(-2)Re^{-2t}\cos (2\sqrt2t-\theta)-Re^{-2t}\sin (2\sqrt2t-\theta)(2\sqrt2)$

Also, $\frac{dx}{dt}=0,\ when\ t=0$ . Putting these values, we get,

$0=(-2)Re^{0}\cos (0-\theta)-Re^{0}\sin (0-\theta)(2\sqrt2) \\ 0=(-2)R\cos (\theta)+R\sin (\theta)(2\sqrt2)$

$0=(-2)0.2+\dfrac{0.2}{\cos\theta}\sin (\theta)(2\sqrt2)$ [Using (ii)]

$1=\tan\theta(\sqrt2)$

$\dfrac1{\sqrt2}=\tan\theta$

$\Rightarrow \theta=35.26\degree$

Put this in (ii)

$R\cos 35.26\degree=0.2$

$R(0.8165)=0.2$

$R=0.2449\approx0.25$

Hence, the position at any time t is: $x(t)=0.25e^{-2t}\cos (2\sqrt2t-35.26\degree)$

(c):

Amplitude$=0.25e^{-2t}\ m$

Period$=\dfrac{2\pi}{2\sqrt2}=\dfrac{\pi}{\sqrt2} \ s$

Frequency$=\dfrac{\sqrt2}{\pi} \ Hz$