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) $4 / 5$

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Answer:

Correct Answer: 5. (a)

Solution:

  1. $y=k t^{2} \Rightarrow \frac{d^{2} y}{d t^{2}}=2 k$

$$ \begin{array}{rlrl} \text { or } & a _y & =2 m / s^{2} \\ T _1 & =2 \pi \sqrt{\frac{l}{g}} \\ \text { and } \quad & \quad T _2 & =2 \pi \sqrt{\frac{l}{g+a _y}} \\ \therefore & \frac{T _1^{2}}{T _2^{2}} & =\frac{g+a _y}{g}=\frac{10+2}{10}=\frac{6}{5} \end{array} $$

$\therefore$ Correct answer is (a).



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