A simple pendulum has time period . The point of suspension is now moved upward according to the relation where is the vertical displacement. The time period now becomes . The ratio is [2005]

Question

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 $\frac{T_{1}^{2}}{T_{2}^{2}}$ is $(g = 10 {m/s}^{2})$ [2005]

Smart Achievers · Accepted Answer

(2)

$Given : y = k t^{2}$

$∴$ $\frac{d y}{d t} = v = 2 k t$ and, $\frac{d^{2} y}{d t^{2}} = a = 2 k$

or $a_{y} = 2 {m/s}^{2}$ $(∵ k = 1 {m/s}^{2} given)$

We know that $T = 2 π \sqrt{\frac{l}{g}}$ or, $T_{1} = 2 π \sqrt{\frac{l}{g}}$

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

$∴$ $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{g + a_{y}}{g}$ $\Rightarrow \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{12}{10} = \frac{6}{5}$

Solution

Q.
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 $\frac{T_{1}^{2}}{T_{2}^{2}}$ is $(g = 10 {m/s}^{2})$ [2005]

1 5/6

2 6/5

3 1

4 4/5

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