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Standard XII

Physics

Intro to Simple Pendulum

T1 is the tim...

Question

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

4/5

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6/5

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5/6

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Solution

The correct option is B 6/5
At the point of suspension, acceleration is,

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

$T = 2 π \sqrt{\frac{l}{g}}$

$T_{1} = 2 π \sqrt{\frac{l}{10}}$ and $T_{1} = 2 π \sqrt{\frac{l}{10 + 2}} = 2 π \sqrt{\frac{l}{12}}$

Hence ${(\frac{T_{1}}{T_{2}})}^{2} = \frac{6}{5}$

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T1 is the time period of a 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 6/5At the point of suspension, acceleration is,a=d2ydt2=2k=2m/s2T=2π√lgT1=2π√l10 and T1=2π√l10+2=2π√l12Hence (T1T2)2=65

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