Question

A bob of simple pendulum of mass $m$ performs SHM with amplitude $A$ and period $T$. Kinetic energy of pendulum at displacement $x = \frac{A}{2}$ will be

• A. $\frac{2\pi^2 mA^3}{T^2}$
• B. $\frac{3mA^2\pi^2}{T}$
• C. $\frac{2\pi mA^2}{3T}$
• D. $\frac{3\pi^2 mA^2}{2T^2}$

Hint:

At half-amplitude ($x = A/2$), the potential energy is always exactly $1/4$ of the total energy because $P.E. \propto x^2$. Consequently, the remaining kinetic energy must always be exactly $3/4$ of the total energy ($E = \frac{2\pi^2 mA^2}{T^2}$). Multiplying $\frac{3}{4} \times \frac{2\pi^2 mA^2}{T^2}$ instantly yields the correct option.

Answer 1

The Correct Option is D

Step 1: Restate the goal.
A pendulum bob of mass $m$ does SHM of amplitude $A$ and period $T$. We want its kinetic energy at the instant its displacement is $x = \tfrac{A}{2}$.
Step 2: Recall total energy and the energy split.
Total energy in SHM is $E = \tfrac{1}{2} m \omega^2 A^2$. At displacement $x$ the kinetic part is $K = \tfrac{1}{2} m \omega^2 (A^2 - x^2)$, because potential energy takes the $x^2$ share.
Step 3: Insert $x = \tfrac{A}{2}$.
Then $A^2 - x^2 = A^2 - \dfrac{A^2}{4} = \dfrac{3A^2}{4}$. So $K = \tfrac{1}{2} m \omega^2 \cdot \dfrac{3A^2}{4} = \dfrac{3}{8} m \omega^2 A^2$.
Step 4: Express $\omega$ through the period.
Since $\omega = \dfrac{2\pi}{T}$, we get $\omega^2 = \dfrac{4\pi^2}{T^2}$.
Step 5: Substitute and simplify.
$K = \dfrac{3}{8} m \cdot \dfrac{4\pi^2}{T^2} \cdot A^2$. The factor $\dfrac{3 \times 4}{8} = \dfrac{3}{2}$, so $K = \dfrac{3\pi^2 m A^2}{2 T^2}$.
Step 6: Match the option.
This is exactly option (4). A neat check: at $x = \tfrac{A}{2}$ the kinetic energy is $\tfrac{3}{4}$ of the total, consistent with potential energy being $\tfrac{1}{4}$ of it.
\[ \boxed{K = \dfrac{3\pi^2 m A^2}{2T^2}} \]