An oscillating pendulum suspended from the roof of a lift which is at rest has time period T₁. When lift moves up with acceleration 'a' its time period is T₂. When lift moves down with acceleration 'a' its time period is T₃. The relation between T₁, T₂ and T₃ is (a) T₁ = 2T₂T₃√T₂²+T₃² (b) T₁ = √(2) T₂T₃√T₂²+T₃² (c) T₁ = 2T₂²T₃²√(T₂+T₃) (d) T₁ = √(2) T₂²T₃²√(T₂+T₃)

Answer: (b). T ∝ 1√geff ⇒ 1/T² ∝ geff. Lift up: g+a; lift down: g-a. Adding: (1)/(T₂²) + (1)/(T₃²) = (2)/(T₁²). T₁² = (2T₂²T₃²)/(T₂²+T₃²). So the answer is (b) T₁ = (√(2) T₂T₃)/(√(T₂²+T₃²)).

An oscillating pendulum suspended from the roof of a lift which is at… | MHT CET 2026 (11 Apr, Shift 2)

Q17MHT CET 2026 · 11 APR · SHIFT 2Medium

Simple Pendulum

An oscillating pendulum suspended from the roof of a lift which is at rest has time period $T_1$ . When lift moves up with acceleration 'a' its time period is $T_2$ . When lift moves down with acceleration 'a' its time period is $T_3$ . The relation between $T_1,\ T_2$ and $T_3$ is

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Answer: (b)

$T \propto \frac{1}{\sqrt{g_{eff}}} \Rightarrow \frac{1}{T^2} \propto g_{eff}$ .
Lift up: $g+a$ ; lift down: $g-a$ . Adding: $\frac{1}{T_2^2} + \frac{1}{T_3^2} = \frac{2}{T_1^2}$ .
$T_1^2 = \frac{2T_2^2T_3^2}{T_2^2+T_3^2}$ .
So the answer is (b) $T_1 = \frac{\sqrt{2}\,T_2T_3}{\sqrt{T_2^2+T_3^2}}$ .

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Simple Pendulum question from MHT CET 2026 (11 Apr, Shift 2)

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