Question

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of 3.27 Hz. (a) What is the spring constant of each spring if the mass of the car is 1140 kg and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging 68.0 kg each, ride in the car with an even distribution of mass?

Answer 1

Answer:

a) 120341 N/m

b) 1.432 Hz

Step-by-step explanation:

Given that

Frequency of oscillations of spring, F = 3.27 Hz

Mass of the car, m = 1140 kg

Average mass of passengers, M = 68 kg

The mass on each spring,

m = 1/4 m,

m = 1140 / 4

m = 285 kg

The spring constant on each spring, k is

ω = √(k/m), also,

ω = 2πf, so that

2πf = √(k/m)

If we square both sides, then

4π²f² = k/m

k = 4π²f²m if we substitute, we have

k = 4 * 3.142² * 3.27² * 285

k = 120341 Nm

Since the mass of each person is 68 kg, then the total mass of all passengers would be

68 * 5 = 340 kg

Thus, the total mass of the oscillating system is,

mass of passengers + mass of car

340 kg + 1140 kg

M(total) = 1480 kg

To get the vibrating frequency, we substitute for values in the equation

2πf = √(k/m)

F = 1/2π * √(k/m)

F = 1/2π * √(120341 / 1480)

F = 1/2π * √81.31

F = 0.1591 * 9

F = 1.432 Hz