A compact car has a mass of 1310 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.

Required:
a. What is the spring constant of each spring if the empty car bounces up and down 2.0 times each second?
b. What will be the carâ€™s oscillation frequency while carrying four 70 kg passengers?

In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.

The expression for the angular velocity is

w = √k/m

the angular velocity is related to the period

w = 2π / T

we substitute

T = 2[tex]\pi[/tex] √m/ k

a) empty car

k = 4π² m / T²

k = 4 π² 1310/2 2

k = 12929.18 N / m

This is the equivalent constant of the short springs

F1 + F2 + F3 + F4 = k_eq x

k x + kx + kx + kx = k_eq x

k_eq = 4 k

k = k_eq / 4

k = 12 929.18 / 4

k= 3232.30 N / m

b) the frequency of oscillation when carrying four passengers.

In this case the plus is the mass of the vehicle plus the masses of the passengers

m_total = 1360 + 4 70

m_total = 1640 kg

angular velocity and frequency are related

w = 2pi f

we substitute

2 pi f = Ra K / m

in this case the spring constant changes us

k_eq = 12929.18 N / m

f = 1 / 2π √ 12929.18 / 1640

f = π / 2 2.80778

f = 4,410 Hz