The angular frequency of the damped oscillator is given by `omega=sqrt((k)/(m)-(r^(2))/(4m^(2)))` ,where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio `r^(2)//(m k)` is 8% ,the change in the time period compared to the undamped oscillator

Options

A. increases by 1%
B. increases by 8%
C. decreases by 1%
D. decreases by 8%

Answer

A

Explanation

Given angular frequency of damped oscillator
`omega = sqrt((K)/(m) - (r^(2))/(4m^(2))) rArr T = (2pi)/(sqrt((K)/(m) -(r^(2))/(4m^(2))))" "...(i)`
Time period for undamped oscillator
`T_(0) = (2pi)/(sqrt((K)/(m)))`
From equation (i)
`T = (2pi)/(sqrt((K)/(m))[1-(r^(2))/(4mK)]^((1)/2))`
`T = T_(0) [1-(r^(2))/(4mK)]^((-1)/(2))`
`(T)/(T_(0)) = 1 + (r^(2))/(8mK)`
`((T-T_(0))/(T_(0))) xx 100 = (1)/(8) [(r^(2))/(mK) xx 100] = (1)/(8) xx 8% = 1% `
Change in time period compared to undamped oscillator in increased by 1%.