Answer to Question #192864 in Differential Equations for Sourav

Question

1. An object of mass 20 kg is pushed on a floor with a force of 40sin 2t N. Given that the 

frictional force is 20 times the velocity and the object starts from rest, determine the 

velocity of the object as a function of time. 

2. Consider an LCR circuit with L = 0.1 H, C = 0.01 F and R = 3.0. Determine the 

electric current in the circuit, given that at t = 0, the charge in the circuit is zero and the 

current is 2 A.

Answer 1

(1) Force applied, $F=40sin(2t)$

Let force of friction $=F_f=m\dfrac{d(20v)}{dt}=20m\dfrac{dv}{dt}$

Net force, $F_{net}=F-F_f=40sin(2t)-20m\dfrac{dv}{dt}$

$ma=40sin(2t)-20m\dfrac{dv}{dt}$

$a=\dfrac{40sin(2t)}{m}-20\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=\dfrac{40sin(2t)}{m}-20\dfrac{dv}{dt}$

$21dv=\dfrac{(40sin(2t))}{m}dt$

$\intop^v_odv=\dfrac{1}{21}\intop^t_o\dfrac{(40sin(2t))}{m}dt$

$v=\dfrac{40}{21\times20\times2}[-cos(2t)+1]$

$v=0.047(1-cos2t)$

(2) Inductance, $L=0.1\space H$

Capacitance, $C=0.01\space F$

Resistance, $R=3\space \Omega$

By Kirchoff's law we have,

$L\dfrac{d^2q}{dt^2}+R\dfrac{dq}{dt}+\dfrac{q}{C}=0$

taking $R/L=2\lambda=30$ and $1/LC=\mu^2=1000$ we get

$\lambda=15$ and $\mu=31.622$

$\dfrac{d^2q}{dt^2}+2\lambda\dfrac{dq}{dt}+\mu^2q=0$

$D^2+2\lambda D+\mu^2=0$

$D^2+30D+1000=0$

Since, $\mu>\lambda$ , the roots of the auxiliary equation are imaginary, i.e,

$D=-\lambda\pm i\alpha$

where, $\alpha^2=\mu^2-\lambda^2$

The solution of the equation is $q=e^{-\lambda t}(c_1\space cos\space \alpha t+c_2\space sin\space \alpha t)$

at t=0, $q=0$

$0=e^0(c_1+0)$

$c_1=0$

differentiating q with respect to time,

$\dfrac{dq}{dt}=-\lambda e^{-\lambda t}(c_1\space cos\space \alpha t+c_2\space sin\space \alpha t)+e^{-\lambda t}\space\alpha(-c_1\space sin\space \alpha t+c_2\space cos\space \alpha t)$

$i(t)=-\lambda e^{-\lambda t}(c_1\space cos\space \alpha t+c_2\space sin\space \alpha t)+e^{-\lambda t}\space\alpha(-c_1\space sin\space \alpha t+c_2\space cos\space \alpha t)$

At t=0, i=2

$2=\alpha (c_2)$

$c_2=\dfrac{2}{\alpha}=\dfrac{2}{27.83}=0.07$

Substituting the values in the above equation,

$i(t)=c_2e^{-\lambda t}(\alpha\cos\alpha t-\lambda\sin\alpha t)$

$i(t)=0.07e^{-15t}(27.83\cos(27.83t)-15\sin(27.83t))$