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The suspension system of a 1600 kg automobile "sags" 8.9 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 48% each cycle. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming each wheel supports 400 kg.

The suspension system of a 1600 kg automobile "sags" 8.9 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 48% each cycle. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming each wheel supports 400 kg.

Grade:12th pass

1 Answers

manish
15 Points
6 years ago

a).

mg = -kx

so,

k = \frac{mg}{x} = \frac{1600 \times 9.8}{0.089} = 50337.07 N/m

b)

we have,

x(t) = x_{m}e^{-\frac{bt}{2m}}cos(\omega t + \phi)

and we can find time period of oscillation,

T = 2\pi \sqrt{\frac{m}{k}}

T = 2\pi \sqrt{\frac{1600}{50337.07}} = 1.120

so we can write,
\frac{x(t+T)}{x(t)} = e^{\frac{-bt}{2m}}

it is also given that the amplitude decreases BY 55% every cycle, so:

x(t+T) = x(t) - 0.48 x(t) = 0.52x(t)

or

\frac{x(t+T)}{x(t)} = 0.52

so we can write,


e^{\frac{-bT}{2m}} = 0.52

we taken t = T, because at time T damping is occur, but at t = t no damping.
log (e^{\frac{-bT}{2m}}) = log(0.52)

\frac{-bT}{2m} = log(0.52)

\frac{bT}{2m} = log(\frac{1}{0.52})

b = \frac{2m}{T}log(\frac{1}{0.52}) = 811.41kg/m

 

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