Question

1. My pickup truck as a weight of 4500 lbs and a bed capacity of 1000 lbs. When I drive along a bumpy road with an empty truck, the truck oscillates with a frequency f = 4 cycles/sec.

What is the spring constant of the suspension? If I add 1000 lbs in the bed of the truck, at what frequency will the truck oscillate?

Answer 1

(1) Let, the spring constant of the suspension be k.

The empty truck oscillates with a frequency f_o = 4 Hz.

So, angular frequency of this oscillation = _o = 2f_o = 2 x 3.14 x 4 rad / s = 25.12 rad / s.

Empty truck has a mass m_o = ( 4500 / 2.205 ) kg = 2040.82 kg.

So, the spring constant of the suspension k = m_oo² = 2040.82 kg x ( 25.12 rad / s )² = 1.3 x 10⁶ N / m.

When 1000 lbs is added in the bed of the truck, total mass of the truck becomes :

m = { ( 4500 + 1000 ) / 2.205 } kg = 2494.33 kg.

Hence, angular frequency of the oscillation is now = ( k / m )

or, = ( 1.3 x 10⁶ / 2494.33 ) rad / s = 22.83 rad / s.

Hence, frequancy of the truck's oscillation is f = / 2 = { 22.83 / ( 2 x 3.14 ) } Hz = 3.635 Hz.