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}

• A. 3 times
• B. 2 times
• C. 9 times
• D. 5 times

Hint:

"Increase in" refers to $\Delta E$, while "becomes" refers to the final value $E'$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The total energy of an oscillating simple pendulum executing Simple Harmonic Motion (SHM) depends on its mass, angular frequency, and amplitude.
Changing the length of the pendulum changes its angular frequency, which in turn alters its total energy.
Step 2: Key Formula or Approach:
Total energy of a particle in SHM is $E = \frac{1}{2} m \omega^2 A^2$.
For a simple pendulum, the angular frequency squared is $\omega^2 = \frac{g}{L}$, where $L$ is the length.
Substituting this gives $E = \frac{1}{2} m \left(\frac{g}{L}\right) A^2$.
This shows that total energy is inversely proportional to length ($E \propto \frac{1}{L}$) when mass, gravity, and linear amplitude are constant.
Step 3: Detailed Explanation:
Let the initial energy be $E_1$ with length $L_1 = L$.
\[ E_1 = \frac{mgA^2}{2L} \] The new length is made $1/3$ of the original: $L_2 = \frac{L}{3}$.
The new total energy $E_2$ is: \[ E_2 = \frac{mgA^2}{2L_2} = \frac{mgA^2}{2(L/3)} \] Bring the 3 to the numerator: \[ E_2 = 3 \cdot \left(\frac{mgA^2}{2L}\right) \] Notice that the term in parentheses is the initial energy $E_1$: \[ E_2 = 3 E_1 \] The new energy is 3 times the original energy.
However, the question asks for the {increase} in its total energy.
Increase in energy $\Delta E = E_{\text{final}} - E_{\text{initial}}$ \[ \Delta E = E_2 - E_1 = 3E_1 - E_1 = 2E_1 \] The increase is 2 times the original energy.
Step 4: Final Answer:
The increase in total energy will be 2 times.