Question

this si a 2 part question84) A shock absorber is designed to quickly damp out the oscillations that a car would otherwise make because it is suspended on springs. (a) Find the period of oscillation of a 1610-kg car that is suspended by springs that make an effective force constant of 5.75×104 N/m. (b) Find the damping constant b that will reduce the amplitude of oscillations of this car by a factor of 5.00 within a time equal to half the period of oscillation.

Answer 1

Given data

*The given mass of the car is m = 1610 kg

*The given effective force constant is k = 5.75 × 10^4 N/m

(a)

The formula for the period of oscillation of a 1610 kg car is given as

`T=2\pi\sqrt[]{(m)/(k)}`

Substitute the known values in the above expression as

`\begin{gathered} T=2*3.14*\sqrt[]{(1610)/(5.75*10^4)} \\ =1.05\text{ s} \end{gathered}`

Hence, the time period of oscillation of a 1610 kg car is T = 1.05 s

(b)

As from the given data, the amplitude of the oscillation of the car decreases by a factor of 5.00. Then, the expression for the amplitude of the oscillation, and the damping constant (b) is calculated as

`A=A_0e^{-(bt)/(2m)}`

Substitute the known values in the above expression as

`\begin{gathered} (A_0)/(5.0)=A_0e^{-(bt)/(2m)} \\ bt=2m\ln (5.0)_{} \\ b((T)/(2))=2m\ln (5.0) \\ b=(4m\ln (5.0))/(T) \\ =(4*1610*\ln (5.0))/(1.05) \\ =9871.2\text{ kg/s} \end{gathered}`

Hence, the damping constant is b = 9871.2 kg/s