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

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

1.	In a damped oscillatory motion a block of mass 200 g is suspended to a spring of force constant 90 N/m in a medium and damping constant is 40 g/s. Find (a) time period of oscillation(b) time taken for its amplitude of oscillation to drop to half to its initial value (c) time taken for its mechanical energy to drop to half of its initial value.
Answer» Solution :MASS m = 200 g = 0.2 kg FORCE constant k = 90 N/m damping constant b= 40 g/s = 0.04 kg/s `sqrt(KM) = sqrt(90xx0.2) = sqrt(18) kg//s` Here `b lt lt sqrt(km)` a) time period `T= 2PI sqrt((m)/(k)) = 2pi sqrt((0.2)/(90)) = 0.3s` b) amplitude `=Ae^(-bt//2m)` Let amplitude is dropped to half of its initial value after the time `T_(1//2)` . Amplitude `A.e^(-b.T_(1//2))/(2m) = A/2` `e^(-b(T_(1//2)))/(2m) = "In" (1/2)` `impliesT_(1//2) = ("In" (2))/(b//2 m) = 2.302xx0.3010xx2m//b` `T_(1//2) =0.693xx(2m)/(b) = 0.693 xx (2xx0.2)/(0.04) =6.93 s` c) Let the energy is dropped to half of its initial value after a time `t_(1//2)` Initial energy `E_0 = 1/2 KA^2` At time `t_(1//2) `, energy `=1/2 E_0 = 1/2 kA^2 e^(-b.t_(1//2))/m = 1/2 (1/2 kA^2)` `e^(-b.t_(1//2))/(m) =1/2 impliest_(1//2) = LN (2) xx m/b = 0.693 xx m/b` `t_(1//2) = 0.693 xx(0.2)/(0.04) =3.46 s`