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Time Period Independent of Amplitude

A simple pend...

Question

A simple pendulum has time period $T_{1} .$ The point of suspension is now moved upward according to equation $y = K t^{2}$ where $K = 2.5 m / s^{2} .$ If new time period is $T_{2}$ then ratio will be

A

2 / 3

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B

5 / 6

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C

6 / 5

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D

3 / 2

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Solution

The correct option is C $6 / 5$
Given,

Displacement,

$y = k t^{2}$

$\frac{d^{2} y}{d t^{2}} = 2 k$

Since, k=1m/s²

$\frac{d^{2} y}{d t^{2}} = a_{y} = 2 m / s^{2}$

$T_{1} = 2 π \sqrt{\frac{l}{g}} . . . . . . . . . (1)$

$a n d$

$T_{2} = 2 π \sqrt{\frac{l}{g + a_{y}}} . . . . . . . . . (2)$

Taking ratios of equation (1) and (2)

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{g + a_{y}}{g} = \frac{10 + 2}{10} = \frac{6}{5}$

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A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upward according to the relation $y = K t^{2}, (K = 1 m / s^{2})$ where y is the vertical displacement. The time period now becomes $T_{2}$ . The ratio of

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