Question

The shock absorbers in an old car with mass 1160 kg are completely worn out. When a 970 N person climbs slowly into the car, it deforms 3.0 cm. The car is now towed down the road (with the person inside). The car hits a bump, and starts oscillating up and down with an amplitude of 6.4 cm.

Model the car and person as a single body on a spring and find the period and frequency of oscillations.

Answer 1

`f = 0.806\,hz`,
`T = 1.241\,s`

Step-by-step explanation:

The problem can be modelled as a vertical mass-spring system exhibiting a simple harmonic motion. The spring constant is:

`k = (970\,N)/(0.03\,m)`

`k = 32333.333\,(N)/(m)`

The angular frequency is:

`\omega = \sqrt{(32333.333\,(N)/(m) )/(1258.879\,kg) }`

`\omega = 5.068\,(rad)/(s)`

The frequency and period of oscillations are, respectively:

`f = (5.068\,(rad)/(s) )/(2\pi)`

`f = 0.806\,hz`

`T = (1)/(0.806\,hz)`

`T = 1.241\,s`