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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? |
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Answer» Solution :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. |
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