For the damped oscillator of Fig. 15.20, m= 250g, k= 85N/m, and b= 70g/s
How long does it take for the mechanical energy to drop to one-half its initial value?

Explanation

From Eq. 15.35 the mechanical energy at time t is `1//2kx_(m)^(2) e^(-bt//m)`
Calculations: The mechanical energy has the value `1//2kx_(m)^(2)` at t=0, Thus, we must find the value of t for which
`(1)/(2) kx_(m)^(2) e^(-bt//m) = (1)/(2) ((1)/(2) kx_(m)^(2))`.
If we divide both sides of this equation by `1//2kx_(m)^(2)` and solve for t as we did above, we find
`t= (-m In (1)/(2))/(b) = (-(0.25kg) (In (1)/(2)))/(0.070kg//s) = 2.5s`.
This is exactly half the time we calculated in (b), or about 7.5 periods of oscillation. Figure 15.21 was drawn to illustrate this sample problem.