A simple pendulum has time period ' T₁ '. The point of suspension is now moved upward according to equation y = kt² where k = 1 m/s². If new time period is ' T₂ ' then T₁²/T₂² will be ( g = 10 m/s²) (a) 2/3 (b) 5/6 (c) 6/5 (d) 3/2

Answer: (c). y = kt² ⇒ a = (d²y)/(dt²) = 2k = 2 m/s² upward. geff = g + a = 12 m/s². (T₁²)/(T₂²) = geffg = 12/10. So the answer is (c) 6/5.

A simple pendulum has time period ' T₁ '. The point of suspension is… | MHT CET 2025 (21 Apr, Shift 1)

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Simple Pendulum

A simple pendulum has time period ' $T_1$ '. The point of suspension is now moved upward according to equation $y = kt^{2}$ where $k = 1\text{ m/s}^{2}$ . If new time period is ' $T_2$ ' then $T_1^{2}/T_2^{2}$ will be ( $g = 10\text{ m/s}^{2}$ )

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Answer: (c)

$y = kt^2 \Rightarrow a = \frac{d^2y}{dt^2} = 2k = 2\text{ m/s}^2$ upward.
$g_{eff} = g + a = 12\text{ m/s}^2$ .
$\frac{T_1^2}{T_2^2} = \frac{g_{eff}}{g} = \frac{12}{10}$ .
So the answer is (c) $6/5$ .

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