The amplitude of damped oscillator becomes half in one minute. The amplitude after 3 minutes will be `1//x` times the original, where x is

Options

A. `3^(2)`
B. `2xx3`
C. `3 xx2^(2)`
D. `2^(3)`

Answer

D

Explanation

For a damped harmonic oscillator, the amplitude,
`A= A_(0)e^(-lamdat) " where " lamda` is the decay constant
Thus when t = 1 minute, `A=(A_(0))/(2)`
`therefore (A_(0))/(2)=A_(0)e^(-lamda xx 1) therefore (1)/(2) = e ^(-lamda) or e^(lamda)=2`
`therefore " For " t =3, A= A_(0)e^(-3lamda)=(A_(0))/(e^(3lamda))=(A_(0))/((e^(lamda))^(3))=(A_(0))/(2^(3))`
But it is given that `A=(A_(0))/(x)`
`therefore x=2^(3)`