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Standard XII

Physics

Pendulum in Non Inertial Frame

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 considered to be x. Then the value of 5x is ___. (Take $g = 10 m / s^{2}$ )

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Solution

$T_{1} = 2 π \sqrt{\frac{l}{g}}$

$T_{2} = 2 π \sqrt{\frac{L}{g + a}}$

$a = \frac{d^{2} y}{d t^{2}} = 2 k = - 2 m s^{- 2}$

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

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

Q. A simple pendulum has time period

T_{1}

. The point of suspension is now moved 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 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}

)

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 = $10 \frac{m}{s^{2}})$

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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 considered to be x. Then the value of 5x is ___. (Take g=10m/s2)

T1=2π√lg T2=2π√Lg+a a=d2ydt2=2k=−2ms−2 T21T22=g+ag=10+210=65

$T_{1} = 2 π \sqrt{\frac{l}{g}}$

$T_{2} = 2 π \sqrt{\frac{L}{g + a}}$

$a = \frac{d^{2} y}{d t^{2}} = 2 k = - 2 m s^{- 2}$

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