equation when $h=5, k=4$ taking $x=0,v=v_0$ at $t=0$ show that the time of maximum displacement is independent of the initial velocity.

Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : 2014

$$ \dfrac {d^2x}{dt^2}=-k^2x-2h\dfrac {dx}{dt} \\ \therefore \dfrac {d^2x}{dt^2}+2h\dfrac {dx}{dt}+k^2x=0 \\ D^2x+2hDx+k^2x=0 \space where \space D=\dfrac d{dt} \\ (D^2+2hD+k^2)k=0 $$

Auxillary Equation is $D^2+2hD+k^2=0 \\ $

$\therefore D=\dfrac {-2\pm\sqrt{4h^2-4k^2}}2\\ D=-h\pm \sqrt{h^2-k^2}\\ \therefore Y_c=c_1e^{(-h+\sqrt{h^2-k^2})t}+c_2e^{(-h-\sqrt{h^2-k^2})t}\\ put \space h=5\space and \space k=4 \\ \therefore Y_c=c_1e^{-2t}+c_2e^{-8t} \\ $

Now it is given that

$ When \space t=0,x=0\space \space\space \therefore Y_c=0\\ \therefore \dfrac {dy_c}{dt}=C_1(e^{-2t}(-2)-e^{-8t}(-8))\\ \therefore v=-2c_1(e^{-2t}-4e^{-8t}) \\ $

Now it is given that when $t=0,v=V_0 \\ $

$\therefore v_0=-2c_1(e^0-4e^0) \\ \therefore v_0=6c_1\\ \therefore c_1=\dfrac {v_0}6\\ $

Therefore v becomes

$\therefore v=-2\times \dfrac {v_0}6(e^{-2t}-4e^{-8t})\\ \therefore v=\dfrac {v_0}3(4e^{-8t}-e^{-2t})=\dfrac {dy_c}{dt} \\ $

for Maximum displacement $\dfrac {dy_c}{dt} =0 \\ $

$\therefore \dfrac {v_0}3(4e^{-8t}-e^{-2t})=0\\ \therefore 4e^{-8t}-e^{-2t}=0\\ e^{-2t}=4e^{-8t}\\ e^{6t}=4\\ 6t=\log 4\\ t=\dfrac 16\log 4 \\ $

$\therefore $ Time of maximum displacement is independent of initial velocity.