Question

At a place, the length of the oscillating simple pendulum is made $1/4$ times keeping amplitude same then the total energy will be ______.

• A. 2 times
• B. 4 times
• C. 8 times
• D. 16 times

Hint:

Shorter pendulums swing much faster (higher $\omega$). Since Energy scales with the SQUARE of frequency ($E \propto \omega^2$), a pendulum that is 4 times shorter oscillates twice as fast, resulting in 4 times the energy!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The length of a simple pendulum is shortened while its swing amplitude is artificially kept exactly the same. We must determine how this affects the total mechanical energy of the oscillation.

Step 2: Detailed Explanation:
The total energy ($E$) of a particle undergoing Simple Harmonic Motion (SHM) is entirely dependent on its mass ($m$), angular frequency ($\omega$), and amplitude ($A$). The formula is:
$E = \frac{1}{2} m \omega^2 A^2$
For a simple pendulum, the angular frequency $\omega$ is defined by its length ($L$) and gravity ($g$):
$\omega = \sqrt{\frac{g}{L}} \implies \omega^2 = \frac{g}{L}$
Substitute this into the energy equation:
$E = \frac{1}{2} m \left( \frac{g}{L} \right) A^2$
The question explicitly states that the amplitude ($A$) is kept the same, and obviously, mass ($m$) and gravity ($g$) are constant.
Therefore, the total energy is strictly inversely proportional to the pendulum's length:
$E \propto \frac{1}{L}$
If the new length ($L'$) is made $1/4$ of the original length ($L' = L/4$), then the new energy ($E'$) becomes:
$E' \propto \frac{1}{L/4}$
$E' \propto 4 \left( \frac{1}{L} \right)$
$E' = 4E$
The total energy increases by a factor of 4.

Step 3: Final Answer:
The total energy will be 4 times, matching option (b).