Question:
A simple pendulum has time period \(T_1\). The point of suspension is now moved upward according to the relation \(y = kt^2\) where \(k = 1\) ms\(^{-2}\). The time period now becomes \(T_2\). The ratio \(\frac{T_1^2}{T_2^2}\) is \((g = 10\) m/s\(^2)\)
A simple pendulum has time period \(T_1\). The point of suspension is now moved upward according to the relation \(y = kt^2\) where \(k = 1\) ms\(^{-2}\). The time period now becomes \(T_2\). The ratio \(\frac{T_1^2}{T_2^2}\) is \((g = 10\) m/s\(^2)\)
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Upward acceleration of support increases effective g, decreases time period.
Updated On: Apr 23, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
When suspension point accelerates upward, effective \(g_{\text{eff}} = g + a\). Here \(y = kt^2 \Rightarrow a = \frac{d^2y}{dt^2} = 2k = 2\) m/s\(^2\).
Step 2: Detailed Explanation:
\(T_1 = 2\pi\sqrt{\frac{l}{g}}\), \(T_2 = 2\pi\sqrt{\frac{l}{g + a}}\).
\(\frac{T_1^2}{T_2^2} = \frac{g + a}{g} = \frac{10 + 2}{10} = \frac{12}{10} = \frac{6}{5}\).
Step 3: Final Answer:
Thus, \(\frac{T_1^2}{T_2^2} = \frac{6}{5}\).
When suspension point accelerates upward, effective \(g_{\text{eff}} = g + a\). Here \(y = kt^2 \Rightarrow a = \frac{d^2y}{dt^2} = 2k = 2\) m/s\(^2\).
Step 2: Detailed Explanation:
\(T_1 = 2\pi\sqrt{\frac{l}{g}}\), \(T_2 = 2\pi\sqrt{\frac{l}{g + a}}\).
\(\frac{T_1^2}{T_2^2} = \frac{g + a}{g} = \frac{10 + 2}{10} = \frac{12}{10} = \frac{6}{5}\).
Step 3: Final Answer:
Thus, \(\frac{T_1^2}{T_2^2} = \frac{6}{5}\).
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