Question

Been on this problem for 2 hours now:

Four identical springs support equally the weight of a 1150kg car.

If a 91kg driver gets in, the car drops 6.8mm . What is k for each spring?

The car driver goes over a speed bump, causing a small vertical oscillation. Find the oscillation period, assuming the springs aren't damped.

Answer 1

The weight of the driver is supported by each spring equally so each spring experiences a force of the weight/4 which is
(91 x 9.81) / 4
= 223.1775 N

The whole car lowers by 6.8 mm and since all springs are at equal heights, each spring drops down by 6.8 mm. Using
F = kx, we get F/x = k which means k = force/ extension. We need to change the units for extension which is measured in metres.
6.8 mm = 0.0068 m.

k = 223.1775/0.0068
= 32820.22 Nm^-1

This second part simple harmonic motion formulae which I'm assuming you know about.

F= ma and F = kx

ma= kx
a = kx/m since a= w^2x
w^2x = kx/m
w^2 = k/m since w= 2π/T
(2π/T)^2 = k/m
2π/T = √(k/m)
T = 2π x √(m/k)

m= 1150 + 91 = 1241 kg
k = 32820.22 Nm^-1

therefore

T = 2π x √(1241/32820.22)
= 1.22 seconds