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The spring of an automobile trailer is compressed by 0.1m under its own weight of 500 kg. find the critical speed, When the trailer is travelling over a road with a profile approx.
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By a sine wave of amp 008 and wave length of 14 meters. What would be the amp if the trailer was to move at 60km/hr.

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$- \delta st$ = static deflection = 0.1m

M = 500kg

Find critical speed.

Case 1: when critical speed is asked, then system will be under resonance.

$\therefore \ r \ = \ 1$

$r \ = \ \frac{w}{W_n} \ = \ 1$

$W \ = \ w_n$

$2 \pi f = 2 \pi fn$

F = fn

$\frac{V_c}{\lambda } = f_n$ - - - -(1)

$w_n = 2 \pi fn$

Also, $W_n = \sqrt{ \frac{9}{\delta st}} = \sqrt{ \frac{9.81}{0.1}} = 9.90$ rad/sec

$\therefore f_n$ = 1.576 Hz

$\therefore$ $\frac{V_c}{\lambda} = 1.576$

$V_c$ = 1.576 x 14

$V_c$ = 22.069 m/sec

Case 2: y = 0.08m

$\lambda$ = 14m

V = 60 km/hr

Amplitude = ?

Amplitude ratio is given by,

$\frac{X}{Y} = \frac{\sqrt{ 1+ (2 \xi r)^2}}{\sqrt{ (1- r^2)^2 + (2 \xi r)^2}}$

$\therefore$ $\frac{x}{y} = \frac{1}{1 – r^2}$

$r = \frac{w}{w_n}$

$W = 2 \pi f$

$f = \frac{v}{\lambda} = \frac{ (60 \times 1000) \times (3600)^{-1}}{14} = 1.190$ Hz

W = 7.479 rad/sec

R = 0.7555

$\therefore$ $x = 0.08 \ [\frac{1}{1-(0.755)^2}]$

X = 0.186m

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