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» <html><body><p>2.5 s<br/>3.5 s<br/>4.5 s<br/>7.5 s</p>Solution :The energy of the damped oscillator at any instant t is given by <br/> `E=E_(0)e^(-bt//m)` <br/> Where `E_(0)` is its initial enegy and b is the damping constant. <br/> At `t=t_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>//2)`, the energy drop to half of its initial value. <br/> From eq. (i), we get <br/> `(E_(0))/(2)=E_(0)e^(-bt_(1//2)/m)""(1)/(2)=e^(-bt_(1//2)//m)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/taking-1238585" style="font-weight:bold;" target="_blank" title="Click to know more about TAKING">TAKING</a> natural logarithm on both <a href="https://interviewquestions.tuteehub.com/tag/sides-1207029" style="font-weight:bold;" target="_blank" title="Click to know more about SIDES">SIDES</a>, we get <br/> `ln((1)/(2))=-(bt_(1//2))/(m),t_(1//2)=-(mln(1//2))/(b)`. . . (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) <br/> here, `ln(1//2)=-0.693` <br/> `b=40gs^(-1),m=200g` <br/> substituting in eq. (ii) , we get <br/> `t_(1//2)=(0.693xx200g)/(40gs^(-1))=3.5s`</body></html>