1. Energy of a simple pendulum for small oscillations

For small angular displacement, a simple pendulum executes SHM. If $d$ is the linear displacement from mean position, then:

• Total energy is constant: $E = \frac{1}{2}kA^2$
• Potential energy at displacement $d$ is: $PE = \frac{1}{2}kd^2$
• Kinetic energy is: $KE = E - PE = \frac{1}{2}k(A^2-d^2)$

Here, $A$ is amplitude.

2. Variation of $PE$ with displacement $d$

Since $PE \propto d^2,$ its graph is an upward opening parabola with:

• minimum at $d=0$
• symmetric about $d=0$
• maximum at $d=\pm A$

3. Variation of $KE$ with displacement $d$

Since $KE \propto A^2-d^2,$ its graph is a downward opening parabola with:

• maximum at $d=0$
• zero at $d=\pm A$
• symmetric about $d=0$

4. Required graph

So the correct schematic must show:

• $PE$: upward parabola centered at $d=0$
• $KE$: downward parabola centered at $d=0$
• both symmetric about the origin in displacement

These two curves intersect at some intermediate values of $d$.

5. Matching with options

The option representing this correctly is D.

6. Comparison with stored answer

Stored correct answer: D

My derived answer: D

Hence, they agree.