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.

Question

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creamydhaka creamydhaka · Answer 1 · 2020-02-06T08:25:56+01:00

Answer:

A) [tex]k=34867.3384\ N.m^{-1}[/tex]

B) [tex]\omega'\approx84\ Hz[/tex]

Explanation:

Given:

mass of car, [tex]m=1380\ kg[/tex]

A)

frequency of spring oscillation, [tex]f=1.6\ Hz[/tex]

We knkow the formula for spring oscillation frequency:

[tex]\omega=2\pi.f[/tex]

[tex]\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f[/tex]

[tex]\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6[/tex]

[tex]k_{eq}=139469.3537\ N.m^{-1}[/tex]

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):

[tex]k=\frac{k_{eq}}{4}[/tex]

[tex]k=\frac{139469.3537}{4}[/tex]

[tex]k=34867.3384\ N.m^{-1}[/tex]

B)

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

[tex]\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }[/tex]

[tex]\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }[/tex]

[tex]\omega'\approx84\ Hz[/tex]