A simple pendulum has time period $T_{1} .$ The point of suspension is now moved upward according to the relation $y = k t^{2},$ where $y$ is the vertical displacement and $k = 1 {ms}^{- 2}$ . The time period now becomes $T_{2}$ . The ratio $\frac{T_{1}^{2}}{T_{2}^{2}}$ is equal to
$(g = 10 {ms}^{- 2})$

\frac{6}{5}

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\frac{5}{6}

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\frac{4}{5}

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Solution

The correct option is A $\frac{6}{5}$
The point of suspension is moved up according to the relation :
$y = k t^{2}$
This implies that the simple pendulum is in an accelerated frame moving up with acceleration.
$a_{y} = \frac{d^{2} y}{d t^{2}} = 2 k$
$a_{y} = 2 {ms}^{- 2}$ $(∵ k = 1 {ms}^{- 2})$
We know that,
$T_{1} = 2 π \sqrt{\frac{l}{g}}$
and
$T_{2} = 2 π \sqrt{\frac{l}{g_{e f f}}} = 2 π \sqrt{\frac{l}{g + a_{y}}}$

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

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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 become

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 $y = k t^{2}, (k = \frac{1 m}{s^{2}})$ 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 = $10 \frac{m}{s^{2}})$