Simple Harmonic Motion 3 Question 5
5. A simple pendulum has time period $T _1$. The point of suspension is now moved upward according to the relation $y=k t^{2},\left(k=1 m / s^{2}\right)$, where $y$ is the vertical displacement.
The time period now becomes $T _2$. The ratio of $\frac{T _1^{2}}{T _2^{2}}$ is
$$ \text { (Take, } g=10 m / s^{2} \text { ) } $$
(2005, 2M)
(a) $6 / 5$
(b) $5 / 6$
(c) 1
(d) $4 / 5$
Show Answer
Answer:
Correct Answer: 5. (a)
Solution:
- $y=k t^{2} \Rightarrow \frac{d^{2} y}{d t^{2}}=2 k$
$$ \begin{array}{rlrl} \text { or } & a _y & =2 m / s^{2} \\ T _1 & =2 \pi \sqrt{\frac{l}{g}} \\ \text { and } \quad & \quad T _2 & =2 \pi \sqrt{\frac{l}{g+a _y}} \\ \therefore & \frac{T _1^{2}}{T _2^{2}} & =\frac{g+a _y}{g}=\frac{10+2}{10}=\frac{6}{5} \end{array} $$
$\therefore$ Correct answer is (a).
