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Q.A simple pendulum is suspended from the roof of a school bus which moves in a horizontal direction with an acceleration 'a', then the time period is :

(a) T is proportional to sqrt(g^2 + a^2)
(b) T is proportional to 1/(g^2 + a^2)
(c) T is proportional to (g^2 + a^2)
(d) T is proportional to 1/sqrt(g^2 + a^2)
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2022MCQ· 1mImportance★★★★★
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A pendulum in a horizontally accelerating vehicle feels an 'effective gravity' that is the vector sum of true gravity g and the pseudo-force per unit mass a: g_eff = √(g²+a²). Its time period is then T = 2π√(L/g_eff), so T decreases as g_eff (and hence as g²+a²) increases.

Work in the non-inertial frame of the bus. Two 'forces' act per unit mass on the bob: real gravity g, straight down; and the pseudo-force a, horizontal, pointing backward relative to the bus's forward acceleration. These two are perpendicular to each other, so they combine (vector sum) to give an effective gravitational field:

g_eff = √(g² + a²)

The pendulum still executes SHM about the new equilibrium direction (tilted, along g_eff instead of straight down), with time period given by the usual simple-pendulum formula but using g_eff in place of g:

T = 2π √(L / g_eff) = 2π √[ L / √(g²+a²) ]

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