Question 233: Oscillations

Question

If the length of the oscillating simple pendulum is made $\frac{1}{3}$ times the original keeping amplitude same then increase in its total energy at a place will be

Accepted Answer

Given:
The length of the pendulum is reduced to $\frac{1}{3}$ of the original.
Let the original length be $l$.
Step 1: Expression for Total Energy
The total energy $E$ of a simple pendulum (for small amplitudes) is given by:
$$ E = \frac{1}{2} m \omega^2 A^2 $$
where
$m$ = mass of bob
$\omega$ = angular frequency
$A$ = amplitude (linear)
Step 2: Angular Frequency for Simple Pendulum
$$ \omega = \sqrt{\frac{g}{l}} $$
Step 3: Calculate Initial and Final Energy
Let $E_1$ = initial energy (length $= l$)
Let $E_2$ = final energy (length $= \frac{l}{3}$)
Initial:
$$ E_1 = \frac{1}{2} m (\sqrt{\frac{g}{l}})^2 A^2 = \frac{1}{2} m \frac{g}{l} A^2 $$
Final:
$$ E_2 = \frac{1}{2} m \left(\sqrt{\frac{g}{l/3}}ight)^2 A^2 = \frac{1}{2} m \left(\frac{g}{l/3}ight) A^2 = \frac{1}{2} m \frac{3g}{l} A^2 $$
Step 4: Find Ratio and Increase
$$ \frac{E_2}{E_1} = \frac{\frac{1}{2} m \frac{3g}{l} A^2}{\frac{1}{2} m \frac{g}{l} A^2} = \frac{3g}{g} = 3 $$
Increase in energy = $E_2 - E_1 = 3 E_1 - E_1 = 2 E_1$
So,
increase is 2 times the original energy.
Correct answer:
$ \boxed{	ext{Option B: 2 times}} $

Answer

3 times

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2 times

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5 times

Oscillations

If the length of the oscillating simple pendulum is made \frac{1}{3} times the original keeping amplitude same then increase in its total energy at a place will be

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