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: Understanding the Question:
The problem asks for the kinetic energy of a simple pendulum executing Simple Harmonic Motion (SHM) when its displacement from the mean position is exactly half of its maximum amplitude ($x = A/2$).

Step 2: Key Formula or Approach:
The kinetic energy ($K.E.$) of a particle performing SHM at any displacement $x$ is given by:
$$K.E. = \frac{1}{2}m\omega^2(A^2 - x^2)$$ The angular frequency $\omega$ is related to the time period $T$ by the equation:
$$\omega = \frac{2\pi}{T}$$

Step 3: Detailed Explanation:
Substitute the given displacement $x = \frac{A}{2}$ into the kinetic energy expression:
$$K.E. = \frac{1}{2}m\omega^2\left(A^2 - \left(\frac{A}{2}\right)^2\right)$$ $$K.E. = \frac{1}{2}m\omega^2\left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2}m\omega^2\left(\frac{3A^2}{4}\right) = \frac{3}{8}m\omega^2A^2$$ Now, substitute the expression for angular frequency $\omega = \frac{2\pi}{T}$:
$$K.E. = \frac{3}{8}m\left(\frac{2\pi}{T}\right)^2A^2$$ $$K.E. = \frac{3}{8}m\left(\frac{4\pi^2}{T^2}\right)A^2$$ Simplifying the fractions ($\frac{3 \times 4}{8} = \frac{3}{2}$):
$$K.E. = \frac{3\pi^2 mA^2}{2T^2}$$

Step 4: Final Answer:
The kinetic energy at $x = \frac{A}{2}$ is $\frac{3\pi^2 mA^2}{2T^2}$, which corresponds to option (D).