**1 Answer**

written 4.3 years ago by |

rail road car and has a stiffness $2 \times 10^5$N/m.

**If the rail road car is travelling at a speed of 72Kmph when engages the bumper, what is the maximum deflection of bumper.**

$\rightarrow$ $\xi = \frac{c}{c_c} = \frac{c}{2mwn} = \frac{c}{2 \times 20 \times 10^3 \times \sqrt{ \frac{2 \times 10^5}{20 \times 10^3}}}$

$c = 1.58 \times 10^5$ N-s/m

$\rightarrow$ $\xi \ \gt \ l$ [over damped system]

$x = c_1 e^{s1t} + c_2 e^{s2t}$

$x = c_1 . e^{-1.581t} + c_2 . e^{-6.324t}$

As, t = 0 , x = 0

$\therefore$ $c_1 = - c_2$

$x = c_1 (-1.581).e^{-1.581t} + c_2 (-6.324).e^{-6.324t}$

As, t = 0 , x = 20m/sec

$20 = c_1 (-1.581) - 6.324 . c_2$

$= -1.581 (-c_2) - 6.324 c_2$

$= 1.581 c_2 - 6.324 c_2$

$20 = -4.743 c_2$

$c_2 = -4.216, c_1 = 4.216$

$\therefore$ $x = 4.216 e^{-1.58t} - 4.216^{-6.324t}$

For max. deflection diff 'x' wrt 't'

$\frac{d}{dt} [ 4.216 e^{-1.581t} - 4.216 e^{-6.324t}] = 0$

$4.216 (-1.581)e^{-1.581t} - 4.216 (-6.324) e ^{-6.324t} = 0$

$- 6.67 e^{-1.581t} + 26.66e^{-6.24t} = 0$

$4 = e^{-(1.581 - 6.24)t}$

$= e^{-(-4.659)t}$

$4 = e^{4.659t}$

1.3862 = 4.659 t

tmax = 0.2975 sec

$x = 4.216 \times e^{-0.470} - 4.216 \times e^{-1.88}$

= 2.632 - 0.643

x = 1.99 m