Question 247: Oscillations

Question

A simple pendulum has time period ' $\mathrm{T}_1$ '. The point of suspension is now moved upward according to equation $\mathrm{y}=\mathrm{kt}^2$ where $\mathrm{k}=1 \mathrm{~m} / \mathrm{s}^2$. If new time period is ' $\mathrm{T}_2$ ' then $\mathrm{T}_1^2 / \mathrm{T}_2^2$ will be ( $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

Step 1: Original Time Period
The time period of a simple pendulum (with no acceleration of suspension point) is
$$ T_1 = 2\pi\sqrt{\frac{l}{g}} $$
Step 2: Acceleration of the Suspension Point
The suspension point moves upward as $y = k t^2$ with $k = 1~	ext{m/s}^2$.
Differentiate twice to get acceleration:
$ y = k t^2 \implies \frac{dy}{dt} = 2kt \implies \frac{d^2y}{dt^2} = 2k $
So, upward acceleration $a = 2k = 2 	imes 1 = 2\ 	ext{m/s}^2$
Step 3: Effective Acceleration Due to Gravity
When the point of suspension is given an upward acceleration $a$, the effective gravity becomes:
$ g' = g + a $
Here,
$ g' = 10 + 2 = 12~	ext{m/s}^2 $
Step 4: New Time Period
The new time period is
$ T_2 = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{12}} $
Step 5: Ratio of the Squares of Time Periods
$ \frac{T_1^2}{T_2^2} = \frac{\left(2\pi\sqrt{\frac{l}{g}}ight)^2}{\left(2\pi\sqrt{\frac{l}{g'}}ight)^2} = \frac{\frac{l}{g}}{\frac{l}{g'}} = \frac{g'}{g} = \frac{12}{10} = \frac{6}{5} $
Final Answer:
Option C
$ \boxed{\frac{6}{5}} $

Answer

$2 / 3$

Answer

$5 / 6$

Answer

$6 / 5$

Answer

$3 / 2$

Oscillations

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