A simple pendulum of length l₁ has time period T₁. Another simple pendulum of length l₂ (l₁ > l₂) has time period T₂. Then the time period of the pendulum of length (l₁ - l₂) will be (a) T₁ - T₂ (b) √((T₁)/(T₂)) (c) √(T₁² - T₂²) (d) √((T₂)/(T₁))

Answer: (c). T² ∝ l: T₁² = kl₁, T₂² = kl₂. For length (l₁ - l₂): T² = k(l₁ - l₂) = T₁² - T₂² So the answer is (c) √(T₁² - T₂²).

A simple pendulum of length l₁ has time period T₁. Another simple… | MHT CET 2024 (4 May, Shift 1)

Q36MHT CET 2024 · 4 MAY · SHIFT 1Medium

Simple Pendulum

A simple pendulum of length $l_1$ has time period $T_1$ . Another simple pendulum of length $l_2\ (l_1 > l_2)$ has time period $T_2$ . Then the time period of the pendulum of length $(l_1 - l_2)$ will be

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Answer: (c)

$T^2 \propto l$ : $T_1^2 = kl_1$ , $T_2^2 = kl_2$ .
For length $(l_1 - l_2)$ : $T^2 = k(l_1 - l_2) = T_1^2 - T_2^2$
So the answer is (c) $\sqrt{T_1^2 - T_2^2}$ .

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Simple Pendulum question from MHT CET 2024 (4 May, Shift 1)

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