Question

A body is performing S.H.M. of amplitude 'A'. The displacement of the body from a point where kinetic energy is maximum to a point where potential energy is maximum, is

• A. zero
• B. $\pm A$
• C. $\pm \frac{A}{2}$
• D. $\pm \frac{A}{4}$

Hint:

In S.H.M., maximum kinetic energy lives entirely at the center ($x=0$), while maximum potential energy lives exclusively at the outer walls ($x = \pm A$). The distance from the center to either wall is by definition the amplitude $A$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the structural displacement covered when moving from the position in simple harmonic motion (S.H.M.) where kinetic energy reaches its peak value to the position where potential energy reaches its peak value.

Step 2: Key Formula or Approach:
In simple harmonic motion, energy oscillates continuously between kinetic and potential forms:
1. Kinetic Energy ($K$) is maximum where speed is highest, which occurs at the mean equilibrium position:
$$x_{K\text{-max}} = 0$$ 2. Potential Energy ($U$) is maximum where the spring compression/stretching is greatest, which occurs at the extreme turning positions:
$$x_{U\text{-max}} = \pm A$$

Step 3: Detailed Explanation:
Let's find the absolute distance vector between these two critical milestone positions. The initial position (maximum kinetic energy point) is the equilibrium reference mark: $x_1 = 0$.
The final position (maximum potential energy point) corresponds to the extreme limits of oscillation: $x_2 = \pm A$.
The total displacement vector $\Delta x$ from the first point to the second point is:
$$\Delta x = x_2 - x_1 = (\pm A) - 0 = \pm A$$ This means the body travels a distance equal to the full amplitude $A$ in either the positive or negative direction.

Step 4: Final Answer:
The displacement between these energy extrema points is $\pm A$, corresponding to option (B).