In the question number 93, the time elapsed for its mechanical energy to drop half of its initial value is

In the question number 93, the time elapsed for its mechanical energy

1.	In the question number 93, the time elapsed for its mechanical energy to drop half of its initial value is
Answer» 2.5 s 3.5 s 4.5 s 7.5 s Solution :The energy of the damped oscillator at any instant t is given by `E=E_(0)e^(-bt//m)` Where `E_(0)` is its initial enegy and b is the damping constant. At `t=t_(1//2)`, the energy drop to half of its initial value. From eq. (i), we get `(E_(0))/(2)=E_(0)e^(-bt_(1//2)/m)""(1)/(2)=e^(-bt_(1//2)//m)` TAKING natural logarithm on both SIDES, we get `ln((1)/(2))=-(bt_(1//2))/(m),t_(1//2)=-(mln(1//2))/(b)`. . . (II) here, `ln(1//2)=-0.693` `b=40gs^(-1),m=200g` substituting in eq. (ii) , we get `t_(1//2)=(0.693xx200g)/(40gs^(-1))=3.5s`