Question

The figure shows a graph of displacement versus time for a particle executing S.H.M. The acceleration of the S.H.O. at the end of time t = (4/3) second is ………. cm.s^–2 

(A) (√3/32)π² 

(B) – (π²/32) 

(C) (π²/32) 

(D) – (√3/32)π²

Answer 1

The correct option (D) – (√3/32)π²  

Explanation:

y = Asinωt

V = (dy/at) = Aωcosωt

a = (dv/dt) = – Aω² sin ωt

from graph, A = 1cm

T = 8sec

∴ ω = (2π/T) = (2π/8) = (π/4)

t = (4/3)

hence a = (– 1) (π/4)² sin {(π/4) × (4/3)}

∴ a = – (π/4)² × sin (π/3)

= – (√3/2) × (π²/16)

= – (√3/32)π²