Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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.

manish
15 Points
4 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,
$\dpi{120} \frac{x(t+T)}{x(t)} = e^{\frac{-bt}{2m}}$

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

$\dpi{120} \frac{x(t+T)}{x(t)} = 0.52$

so we can write,

$\dpi{120} e^{\frac{-bT}{2m}} = 0.52$

we taken t = T, because at time T damping is occur, but at t = t no damping.
$\dpi{120} log (e^{\frac{-bT}{2m}}) = log(0.52)$

$\dpi{120} \frac{-bT}{2m} = log(0.52)$

$\dpi{120} \frac{bT}{2m} = log(\frac{1}{0.52})$

$\dpi{120} b = \frac{2m}{T}log(\frac{1}{0.52}) = 811.41kg/m$