For the damped oscillatore shown in figure., the mass m of the block is 200g, `K=90Nm^(-1)` and the damping constant b is `40gs^(-1)`. Calculate (a) the period of oscillation, (b) time taken for its amplitude of vibrations to drop to half of its initial value of (c) the itme taken for its mechanical energy to drop to half its inital value.

Explanation

(a) Here, `m=200g=0.2kg, `
`k=90Nm^(-1)`
`:. (k)/(m)=(90)/(0.2)=450Nm^(-1)kg^(-1)`
`(b^(2))/(4m^(2))=((0.04)^(2))/(4xx(0.2)^(2))=0.01kg^(2)s^(-2)` ,brgt As, `(k)/(m)gt gt(b^(2))/(4m^(2))` ,brgt Therefore, `omega^(')=sqrt((k)/(m))-(b^(2))/(4m^(2))~~sqrt((k)/(m))`
`T=(2pi)/(omega^('))=2pisqrt((m)/(k))=2xx(22)/(7)sqrt((0.2)/(90))~~0.3s`
(b) If T is the time, when amplitude drops to half value then amplitude of the sampled oscillations at time t is
`a=x_(0)e^(-bt//2m)when t=T,a=x_(0)//2`
`:.(x_(0))/(2)=x_(0)e^(-bt//2m) or 2=e^(bT//2M)`
or `log_(e)2=(bT)/(2m)`
or `T=(2m)/(b)log_(e)2=(2.3026xx2m)/(b)log_(10)2`
`=(2.3026xx2xx0.2xx0.3010)/(0.04)=6.93s`
(c) If `T^(')` is the tiem, when mechanical energy drops to half its mechanical energy `E_(0)`, then mechanical energy E of the dampled oscillations at an instant t is given by `E=E_(0)e^(-bt//m)`
When `t=T^('), E=E_(0)//2, ` then
`(E_(0))/(2) =E_(0)e^(-bT^(')//m)or 2=c^(bT^(')//m)`
or `log_(e)2=(bT^('))/(m)`
or `T^(')=2.3026log_(10)2xx(m)/(b)`
or `T^(')=2.3026xx0.3010xx(0.2)/(0.04)=3.46s`