Simple Harmonic Motion Ques 27
- 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: 27.(a)
Solution:
Formula:
- $y=k t^{2} \Rightarrow \frac{d^{2} y}{d t^{2}}=2 k$
$\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}} $
$\therefore \frac{T _1^{2}}{T _2^{2}} =\frac{g+a _y}{g}=\frac{10+2}{10}=\frac{6}{5}$
$\therefore$ Correct answer is (a).
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