A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upwards 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 $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{x}{10}$ . Find the value of x $(g = 10 m / s^{2})$

Open in App

Solution

Since, $y = k t^{2}$
$∴ \frac{d y}{d t} = 2 k t$
$\frac{d^{2} y}{d t^{2}} = 2 k . . . (i) (∵ k = 1 m / s^{2} given)$
$a = 2 m / s^{2}$
We know that $T = 2 π \sqrt{\frac{l}{g}}$
$∴ \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{g_{2}}{g_{1}}$
As the point of suspension accelerate in upward direction, a Pseudo force acts in opposite of the the motion's direction.
Therefore, here $g_{2}$ is effective acceleration due to gravity because of motion of point of suspension.
$[∵ g_{1} = 10 m / s^{2} g_{2} = g + 2 = 12 m / s^{2}]$
$\Rightarrow \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{12}{10} = \frac{6}{5} = \frac{x}{10}$
$∴ x = 12$

Suggest Corrections

Similar questions

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 of

\frac{T_{1}^{2}}{T_{2}^{2}}

(g = 10 m / s^{2})

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 of

\frac{T_{1}^{2}}{T_{2}^{2}}

is (Take

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 $y = k t^{2} (k = 1 m / s^{2})$ , where y is the vertical displacement. The time period now becomes $T_{2}$ . The ratio of $\frac{T_{1}^{2}}{T_{2}^{2}}$ is considered to be x. Then the value of 5x is ___. (Take $g = 10 m / s^{2}$ )

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