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 amplitude of oscillation to drop to half of its initial value

In damped oscillatory motion a block of mass 200g is suspended to a sp

1.	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 amplitude of oscillation to drop to half of its initial value
Answer» <html><body><p></p>Solution :Mass ,= <a href="https://interviewquestions.tuteehub.com/tag/200g-290204" style="font-weight:bold;" target="_blank" title="Click to know more about 200G">200G</a>= 0.2 kg force constant k = `90N//m` damping constant `b = 40g//s =0.04 kg//s`. <br/> `sqrt(km)= sqrt(<a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</a> xx 0.2)= sqrt(18) kg//s` <br/> Here `b lt lt sqrt(km)`<br/> amplitude = `Ae^(-bt//2m)` <br/> Let amplitude is dropped to half of its initial value after the time `T_(1//2)`: <br/> Amplitude `Ae^((-bT_(1//2))/(2m))= (A)/(2) implies e^((-bt_(1//2))/(2m))= (1)/(2)` <br/> Take <a href="https://interviewquestions.tuteehub.com/tag/natural-575613" style="font-weight:bold;" target="_blank" title="Click to know more about NATURAL">NATURAL</a> logarithms on both sides <br/> `(-bT_(1//2))/(2m)= "<a href="https://interviewquestions.tuteehub.com/tag/ln-1076444" style="font-weight:bold;" target="_blank" title="Click to know more about LN">LN</a>"((1)/(2))implies T_(1//2)= ("ln"(2))/(b//2m)= 2.302 xx 0.3010xx2m//b` <br/> `T_(1//2)= 0.693 xx(2m)/(b)= 0.693xx (2 xx 0.2)/(0.04)= 6.93s`</body></html>