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Standard XII

Physics

Variation of G Due to Height and Depth

A simple pend...

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 of $\frac{T_{1}^{2}}{T_{2}^{2}}$ is $(g = 10 m / s^{2})$ :

\frac{6}{5}

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\frac{5}{6}

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1

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\frac{4}{5}

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Solution

The correct option is A $\frac{6}{5}$
$y = K t^{2} y = t^{2} \frac{d y}{d t} = 2 t \frac{d^{2} y}{{d t}^{2}} = 2 = a y$
Effective gravity $= (g + 2) = 12 m / s^{2}$
$T_{1} = 2 π \sqrt{\frac{l}{g}} T_{2} = 2 π \sqrt{\frac{l}{g + 2}} \frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{12}{10} = \frac{6}{5}$

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Similar questions

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

Q. 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})

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}

)

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},

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})

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

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A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y=Kt2,(K=1m/s2) where y is the vertical displacement. The time period now becomes T2. The ratio of T21T22 is (g=10m/s2):

The correct option is A 65y=Kt2y=t2dydt=2td2ydt2=2=ayEffective gravity=(g+2)=12m/s2T1=2π√lgT2=2π√lg+2T12T22=1210=65

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

The correct option is A $\frac{6}{5}$
$y = K t^{2} y = t^{2} \frac{d y}{d t} = 2 t \frac{d^{2} y}{{d t}^{2}} = 2 = a y$
Effective gravity $= (g + 2) = 12 m / s^{2}$
$T_{1} = 2 π \sqrt{\frac{l}{g}} T_{2} = 2 π \sqrt{\frac{l}{g + 2}} \frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{12}{10} = \frac{6}{5}$