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

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

1.	In demaped oscillatory motion a block of mass 200 g is suspended to a spring of force constant 90N//m in a medium and damping constant is 40g/s.Find a time period of oscillation b. time taken for its amplitude of oscillationto drop to half of its initial value (c) time taken for its mecahnical enrgy to drop to half of its initial value.
Answer» SOLUTION :Mass `m =200g=0.2kg` force constant k=90N/m damping constnat b=40g/s=0.04kg/s `sqrt(km)=sqrt(90xx0.2)=sqrt(18)` kg/s Here `b lt lt sqrt(km)` a. time period `T=2pisqrt(m/k)=2pisqrt(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` `impliese^((-b.T_(1//2))/(2m))=1/2` Take natural lograithms on both sides `(-b.(T_(1//2)))/(2m)="In"(1/2)impliesT_(1//2)=(In(2))//(b//2m)` `=2.302xx0.3010xx2m//b` `T_(1//2)=0.693xx(2m)/b=0.693xx(2xx0.2)/(0.04)=6.93s` c. Let the energyis dropped to half of its initial value after a time `t_(1//2)`. Initial energy `E_(0)=1/2kA^(2)` At time `t_(1//2)`, energy `=1/2E_(0)` `=1/2kA^(2)e^((-b.t_(1//2))/m)=1/2(1/2kA^(2))` `e^((-b.t_(1//2))/m)=1/2impliest_(1//2)=In(2)xxm/b=0.693xxm/b` `t_(1//2)=0.693xx0.2/0.004=3.46s`