A particle executes S.H.M. of amplitude A. If I1, and T2, are the times taken by the particle to traverse from 0 to A/2  and from A I 2 to A respectively, then(T1 / T2) will be equal to

For S.H.M.x=Asin⁡ωt=Asin⁡(2πt/T) he time taken by the particle for travelling from x = 0 to x x =Al2isTt.Hence A2=Asin⁡2πT1T or sin⁡2πT1T=12 or 2πT1T=π2  or  T1=T12 Similarly, time taken bythe particle to traverse from x =0 to x =,4 i.e., one-fourth cycle is T4=T1+T2 T2=T4−T1=T4−T12=T6  From eqs. (1) and (2), we get  T1T2=(T/12)T/6=12