Question

A 17.6-kg object oscillates at the end of a horizontal spring that has a spring constant of 20.0 kN/m. The effect of air resistance is represented by the damping coefficient b = 6.69 kg/s. Find the time for the amplitude of the oscillation to drop to 5.98% of its original value.

Answer 1

The time for the amplitude of the oscillation is 14.82 sec.

Step-by-step explanation:

Given that,

Mass of object = 17.6 kg

Spring constant k= 20.0 kN/m

Damping coefficient b= 6.69 kg/s

Amplitude A= 5.98%

We need to calculate the time for the amplitude of the oscillation

Amplitude at any time is given as

`A_(t)=A_(0)e^{(-bt)/(2m)}`

Put the value into the formula

`0.0598 A_(0)=A_(0)e^{(-6.69* t)/(2*17.6)}`

`0.0598=e^{(-6.69* t)/(2*17.6)}`

`t=14.82\ sec`

Hence, The time for the amplitude of the oscillation is 14.82 sec.