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.

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 mechanical energy 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)` Let the ENERGY is dropped to HALF of its initial value after a time `t_(1//2)` intial energy `E_(0)= (1)/(2)KA^(2)` 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))` `e^((-bt_(1//2))/(m))= (1)/(2) implies t_(1//2)= LN(2) xx (m)/(b) = 0.693 xx (m)/(b)` `t_(1//2)= 0.693 xx (0.2)/(0.04)= 3.46s`