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Standard XII

Physics

Physical Pendulum

T1 is the tim...

Question

$T_{1}$ is the time period of 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 $\frac{T_{1}^{2}}{T_{2}^{2}} = ? (g = 10 m / s^{2})$ .

4 / 5

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Solution

The correct option is B $6 / 5$
original time period is $T_{1} = 2 π \sqrt{l} g$
Now, when going up.
$g^{'} = g + f = 10 + 2 = 12$ $m / s^{2}$
So, $T_{2} = 2 π \sqrt{\frac{l}{g^{'}}}$
$∴ \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{g^{'}}{g} = \frac{12}{10} = \frac{6}{5}$

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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})

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 become

T_{2}

. The ratio of

\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 = l m / s^{2}

. If new time period is

t^{2}

then ratio

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

will be

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

The correct option is B 6/5original time period is T1=2π√lgNow, when going up.g′=g+f=10+2=12 m/s2So, T2=2π√lg′∴T21T22=g′g=1210=65

$T_{1}$ is the time period of 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 $\frac{T_{1}^{2}}{T_{2}^{2}} = ? (g = 10 m / s^{2})$ .

The correct option is B $6 / 5$
original time period is $T_{1} = 2 π \sqrt{l} g$
Now, when going up.
$g^{'} = g + f = 10 + 2 = 12$ $m / s^{2}$
So, $T_{2} = 2 π \sqrt{\frac{l}{g^{'}}}$
$∴ \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{g^{'}}{g} = \frac{12}{10} = \frac{6}{5}$