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» Solution :Mass ,= 200G= 0.2 kg force constant k = `90N//m` damping constant `b = 40g//s =0.04 kg//s`. `sqrt(km)= sqrt(90 xx 0.2)= sqrt(18) kg//s` Here `b lt lt sqrt(km)` amplitude = `Ae^(-bt//2m)` Let amplitude is dropped to half of its initial value after the time `T_(1//2)`: Amplitude `Ae^((-bT_(1//2))/(2m))= (A)/(2) implies e^((-bt_(1//2))/(2m))= (1)/(2)` Take NATURAL logarithms on both sides `(-bT_(1//2))/(2m)= "LN"((1)/(2))implies T_(1//2)= ("ln"(2))/(b//2m)= 2.302 xx 0.3010xx2m//b` `T_(1//2)= 0.693 xx(2m)/(b)= 0.693xx (2 xx 0.2)/(0.04)= 6.93s`