A particle performs linear SHM. At a particular instant, the velocity of the particle is u and acceleration is α (both having the same direction). At another instant velocity is v and acceleration is β 0 < α < β (both in opposite direction to each other).The distance between the two positions is

Step-by-step solution

Let distance be x 1 when velocity is u and acceleration α . Let distance be x 2 when velocity is v and acceleration β . If ω is the angular frequency then α = ω 2 x 1 And β = ω 2 x 2 ∴ α + β = ω 2 ( x 1 + x 2 ) ....... (i) Also u 2 = ω 2 A 2 - ω 2 x 1 2 And v 2 = ω 2 A 2 - ω 2 x 2 2 v 2 - u 2 = ω 2 ( x 1 2 - x 2 2 ) v 2 - u 2 = ω 2 ( x 1 - x 2 ) ( x 1 + x 2 ) ........... (ii) By Equation. (i) we get v 2 - u 2 = ( x 1 - x 2 ) ( α + β ) ∴ x 1 - x 2 = v 2 - u 2 α + β Or x 2 - x 1 = u 2 - v 2 α + β