Question Number 84323 by Rio Michael last updated on 11/Mar/20
![A particle moving in a straight line OX has a displacement x from O at time t where x satisfies the equation (d^2 x/(dt^2 )) + 2(dx/dt) + 3x = 0 the damping factor for the motion is [A] e^(−1) [B] e^(−2t) [C] e^(−3t) [D] e^(−5t)](Q84323.png)
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:{OX}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{displacement}\:{x}\:\mathrm{from}\:{O}\:\mathrm{at}\:\mathrm{time}\:{t}\:\mathrm{where}\:{x}\:\mathrm{satisfies} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} \:}\:+\:\mathrm{2}\frac{{dx}}{{dt}}\:+\:\mathrm{3}{x}\:=\:\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{damping}\:\mathrm{factor}\:\mathrm{for}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{is} \\ $$$$\left[\mathrm{A}\right]\:{e}^{−\mathrm{1}} \\ $$$$\left[\mathrm{B}\right]\:{e}^{−\mathrm{2}{t}} \\ $$$$\left[\mathrm{C}\right]\:{e}^{−\mathrm{3}{t}} \\ $$$$\left[\mathrm{D}\right]\:{e}^{−\mathrm{5}{t}} \\ $$
Answered by TANMAY PANACEA last updated on 11/Mar/20
$${m}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+{b}\frac{{dx}}{{dt}}+{kx}=\mathrm{0}\:\:\:{m}=\mathrm{1}\:\:\:{b}=\mathrm{2}\:\:\:{k}=\mathrm{3} \\ $$$${damping}\:{factor}={e}^{\frac{−{bt}}{\mathrm{2}{m}}} \:={e}^{−\frac{−\mathrm{2}{t}}{\mathrm{2}×\mathrm{1}}} ={e}^{−{t}} \\ $$$$ \\ $$