Question

A simple pendulum has time period $T_{1}$. The point of suspension is now moved upward according to the relation $y=k t^{2}\left(k=1 \,ms ^{-2}\right)$ 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 $g=10 \,ms ^{-2}$ )

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

Answer 1

Answer: (A)

According to the relation, $y=k t^{2} $
$\Rightarrow \frac{d y}{d t}=2 k t$
$\therefore a=\frac{d^{2} y}{d t^{2}}=2 k=2 \times 1=2 $
$[\because k=1]$
The point of suspension is moving upward with acceleration a, then effective acceleration due to gravity on pendulum
$g^{'}=g+a=10+2=12 m / s ^{2} $
$\therefore T_{1}=2 \pi \sqrt{\frac{l}{g}}$
And $T_{2}=2 \pi \sqrt{\frac{l}{g+a}} $
$\therefore \frac{T_{1}^{2}}{T_{2}^{2}}=\frac{g+a}{g}$
$=\frac{12}{10}=\frac{6}{5}$