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: Recall energy positions in S.H.M.
Kinetic energy is largest where the body moves fastest, and potential energy is largest where it momentarily stops.
Step 2: Locate maximum kinetic energy.
Speed is greatest at the mean position, so $K$ is maximum at $x = 0$.
Step 3: Locate maximum potential energy.
The body stops at the extreme positions, so $U$ is maximum at $x = \pm A$.
Step 4: Identify the two points.
Start point $x_1 = 0$ (mean), end point $x_2 = \pm A$ (extreme).
Step 5: Find the displacement between them.
Displacement $= x_2 - x_1 = (\pm A) - 0 = \pm A$.
Step 6: Conclude.
The body shifts by the full amplitude in either direction, so the answer is $\pm A$, option (2).
\[ \boxed{\Delta x = \pm A} \]