A body executes simple harmonic motion. The potential energy (P.E), the kinetic energy (K.E) and total energy (T.E) are measured as a function of displacement x. Which of the following statements is true ?

K.E is maximum when x is maximum

Simple Harmonic Motion PYQ — JEE Main 2003 · Shift 0 · Q161 with Solution

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Correct answer: A

For a body executing simple harmonic motion with amplitude $A$ and spring constant $k$ :

Potential energy: $P.E = \frac{1}{2}kx^2$
Total energy: $T.E = \frac{1}{2}kA^2$ which is constant.
Kinetic energy: $K.E = T.E - P.E = \frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}k(A^2-x^2)$

From $K.E = \frac{1}{2}k(A^2-x^2)$ we see that $K.E$ is maximum when $x^2$ is minimum.

The minimum value of $x^2$ is at $x=0$ So, kinetic energy is maximum at the mean position.

Hence, Option A is true.

Option B: $T.E$ is zero when $x=0$

This is false because total energy is constant: $T.E = \frac{1}{2}kA^2 \neq 0$
Option C: $K.E$ is maximum when $x$ is maximum

At extreme position $x=\pm A$ , $K.E = \frac{1}{2}k(A^2-A^2)=0$ So this is false.
Option D: $P.E$ is maximum when $x=0$

Since $P.E = \frac{1}{2}kx^2$ at $x=0$ , $P.E=0$ which is minimum, not maximum. So this is false.

Therefore, the correct statement is: $\boxed{\text{A: } K.E \text{ is maximum when } x=0}$

Stored correct answer: A

My derived answer: A

So, they agree.