Question

The time period of a simple pendulum is $T_{1}$ when its point of suspension is at rest with respect to the ground. When the point of suspension moves vertically upwards according to the law $y=kt^{2}$ where $k=1 \, ms^{- 2}$ , its new time period is $T_{2}$ , then the value of $\frac{T_{1}^{2}}{T_{2}^{2}}$ is $\left[\right.$ Take $g=10 \, ms^{- 2}\left]\right.$

• A. $\frac{4}{5}$
• B. $\frac{6}{5}$
• C. $\frac{5}{6}$
• D. $1$

Answer 1

Answer: (B)

Acceleration of the point of suspension
$a=\frac{d^{2} y}{d t^{2}}=2k=2m / s^{2}$
$T=2\pi \sqrt{\frac{L}{g_{e f f}}} \, \Rightarrow T_{1}=2\pi \sqrt{\frac{L}{\text{g}}} \, \, $
$T_{2}=2\pi \sqrt{\frac{L}{g + 2}}$
$\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{g + 2}{g}=\frac{10 + 2}{10}=\frac{6}{5}$