Sarthaks Test
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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

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Best answer

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

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