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…

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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

MARKS Content Team · Accepted Answer

Correct answer: (A) (u^2-v^2)/(α+β). 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^2A^2-ω^2x12 And v^2=ω^2 A^2-ω^2 x22 v^2-u^2=ω^2(x12-x22) 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)

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