For particle in SHM, the displacement x of the particle as a function o time t is given as `x=A sin (2pi t)`. Here x is in cm and t is in second. Let the time taken by the particle to travel from x=0 to `x=(A)/(2)` be `t_(1)` and the time taken to travel from `x=(A)/(2)` to x=A be `t_(2)` find `(t_(1))/(t_(2))`

Explanation

Here x=0 at t=0
Also `omega =(2pi)/(T)= 2pi, " ":. T=1s`
At `t=t_(1), x=(A)/(2)`
`(A)/(2)=A sin (2pi t_(1))`
(or) `(1)/(2)= sin (2pi t_(1))`
`:. 2pi t_(1)=(pi)/(6) (or) t_(1)=(1)/(12) s`
Time taken from x =0 to `x=A` is `(T)/(4)=(1)/(4)s`
(or) `t_(1)+t_(2)=(T)/(2)=(1)/(4)s`
Hence `(t_(1))/(t_(2))=(1//12)/(1//6)=(1)/(2)`