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In damped oscillatory motion a block of mass 200g is suspended to a spring of force constant 90 N//m in a medium and damping constant is 40g//s. Find time taken for its mechanical energy to drop to half of its initial value.

Answer» <html><body><p></p>Solution :Mass ,= 200g= 0.2 kg force constant k = `90N//m` <a href="https://interviewquestions.tuteehub.com/tag/damping-943615" style="font-weight:bold;" target="_blank" title="Click to know more about DAMPING">DAMPING</a> constant `b = 40g//s =0.04 kg//s`. <br/> `sqrt(km)= sqrt(90 xx 0.2)= sqrt(18) kg//s` <br/> Here `b lt lt sqrt(km)`<br/> Let the <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> is dropped to <a href="https://interviewquestions.tuteehub.com/tag/half-1014510" style="font-weight:bold;" target="_blank" title="Click to know more about HALF">HALF</a> of its initial value after a time `t_(1//2)` <br/> intial energy `E_(0)= (1)/(2)<a href="https://interviewquestions.tuteehub.com/tag/ka-528931" style="font-weight:bold;" target="_blank" title="Click to know more about KA">KA</a>^(2)`<br/> At time `t_(1//2)` , energy =`(1)/(2)E_(0)= (1)/(2)kA^(2)e^((-bt_(1//2))/(m))= (1)/(2)((1)/(2)kA^(2))` <br/> `e^((-bt_(1//2))/(m))= (1)/(2) implies t_(1//2)= <a href="https://interviewquestions.tuteehub.com/tag/ln-1076444" style="font-weight:bold;" target="_blank" title="Click to know more about LN">LN</a>(2) xx (m)/(b) = 0.693 xx (m)/(b)`<br/> `t_(1//2)= 0.693 xx (0.2)/(0.04)= 3.46s`</body></html>


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