A particle executes SHM in accordance with `x=A"sin"omegat`. If `t_(1)` is the time taken by it to reach from x=0 to `x=sqrt(3)(A//2)`and `t_(2)` is the time taken by it to reach from `x=sqrt(3)//2`A to x=A , the value of `t_(1)//t_(2)` is

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))

A particle executes SHM of amplitude A. If T_1 and T_2 are the times taken by the particle to traverse from 0 to A/2 and from A/2 to A respectively, then T_1/T_2 will be equal to

A particle executes S.H.M. between x = -A and x = + A . The time taken for it to go from 0 to A/2 is T_(1) and to go from A/2 to A is T_(2) . Then

The displacement of a particle undergoing S.H.M. as a function of time t is given by x = A sin((4pit)/(3)) Where x is in cm and t in second The time taken by the particle to travel from (a) x = 0 to A/2 is t_(a) and time taken by it to travel from (b) x = A/2 to A is t_(b) . Calculate the ratio t_(b)//t_(a) .

A particle performs SHM about x=0 such that at t=0 it is at x=0 and moving towards positive extreme. The time taken by it to go from x=0 to x=(A)/(2) is____ time the time taken to go from x=(A)/(2) to A . The most suitable option for the blank space is

A particle performs SHM about x=0 such that at t=0 it is at x=0 and moving towards positive extreme. The time taken by it to go from x=0 to x=(A)/(2) is time the time taken to go from x=(A)/(2) to a. The most suitable option for the blank space is