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Standard XII

Physics

Intro to Simple Pendulum

T1 is the tim...

Question

$T_{1}$ is the time period of simple pendulum. The point of suspension moves vertically upward according to $y = k t^{2}$ , where $k = 1 m / s^{2}$ . New time period is $T_{2}$ , then $T_{2}$ , then $\frac{T_{1}^{2}}{T_{2}^{2}} =$ ? $(g = 10 m / s^{2})$

4 / 5

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6 / 5

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5 / 6

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Solution

The correct option is A $6 / 5$
Acceleration of the point of suspension
$a = \frac{d^{2} y}{d t^{2}} = 2 k = 2 m / s^{2}$
$T = 2 π \sqrt{\frac{L}{g_{e f f}}} \Rightarrow T_{1} = 2 π \sqrt{\frac{L}{10}}$ and
$T_{2} = 2 π \sqrt{\frac{L}{12}}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{6}{5}$

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Similar questions

T_{1}

is the time period of a simple pendulum. The point of suspension moves vertically upwards according to

y = k t^{2}

where

k = 1 m / s^{2}

. New time period is

T_{2}

, then

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

Q. A simple pendulum has time period

T_{1} .

The point of suspension is now moved upward according to equation

y = K t^{2}

where

K = 2.5 m / s^{2} .

If new time period is

T_{2}

then ratio will be

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 equation y =

K t^{2}

where K = 1

m / s^{2}

. If new time period is

T_{2}

then find ratio of

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and

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T1 is the time period of simple pendulum. The point of suspension moves vertically upward according to y=kt2, where k=1m/s2. New time period is T2, then T2, then T21T22= ? (g=10m/s2)

The correct option is A 6/5Acceleration of the point of suspensiona=d2ydt2=2k=2m/s2T=2π√Lgeff⇒T1=2π√L10 andT2=2π√L12T21T22=65

$T_{1}$ is the time period of simple pendulum. The point of suspension moves vertically upward according to $y = k t^{2}$ , where $k = 1 m / s^{2}$ . New time period is $T_{2}$ , then $T_{2}$ , then $\frac{T_{1}^{2}}{T_{2}^{2}} =$ ? $(g = 10 m / s^{2})$

The correct option is A $6 / 5$
Acceleration of the point of suspension
$a = \frac{d^{2} y}{d t^{2}} = 2 k = 2 m / s^{2}$
$T = 2 π \sqrt{\frac{L}{g_{e f f}}} \Rightarrow T_{1} = 2 π \sqrt{\frac{L}{10}}$ and
$T_{2} = 2 π \sqrt{\frac{L}{12}}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{6}{5}$