Question

The springs of a 1500 kg car compress 5.00 mm when its 68 kg driver gets into the driver's seat. Part A If the car goes over a bump, what will be the frequency of oscillations

Answer 1

Answer:

the frequency of the oscillation is 1.5 Hz

Explanation:

Given;

mass of the spring, m = 1500 kg

extention of the spring, x = 5 mm = 5 x 10⁻³ m

mass of the driver = 68 kg

The weight of the driver is calculated as;

F = mg

F = 68 x 9.8 = 666.4 N

The spring constant, k, is calculated as;

k = F/m

k = (666.4 N) / (5 x 10⁻³ m)

k = 133,280 N/m

The angular speed of the spring is calculated;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{133280}{1500} } = 9.426 \ rad/s$

The frequency of the oscillation is calculated as;

ω = 2πf

f = ω / 2π

f = (9.426) / (2π)

f = 1.5 Hz

Therefore, the frequency of the oscillation is 1.5 Hz