Question

A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs.
Part A
What is the spring constant of each spring if the empty car bounces up and down 1.6 times each second?
Express your answer using two significant figures.in N/m.

Part B
What will be the car's oscillation frequency while carrying four 70 kg passengers?
Express in two sig figs in Hz.

Answer 1

Answer:

A) $k=34867.3384\ N.m^{-1}$

B) $\omega'\approx84\ Hz$

Explanation:

Given:

mass of car, $m=1380\ kg$

A)

frequency of spring oscillation, $f=1.6\ Hz$

We knkow the formula for spring oscillation frequency:

$\omega=2\pi.f$

$\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f$

$\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6$

$k_{eq}=139469.3537\ N.m^{-1}$

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

So, the stiffness of each spring is (as they are identical):

$k=\frac{k_{eq}}{4}$

$k=\frac{139469.3537}{4}$

$k=34867.3384\ N.m^{-1}$

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

$\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }$

$\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }$

$\omega'\approx84\ Hz$