Question

In simple harmonic motion, at mean position

• A. KE maximum
• B. PE maximum
• C. KE zero
• D. Acceleration maximum

Hint:

Mean position means zero displacement, so zero potential energy and maximum kinetic energy!

Answer 1

The Correct Option is A

Step 1: What is the mean position?
In SHM the mean position is the centre, where the displacement is zero ($x=0$). The body just passes through it.

Step 2: Speed at the centre.
The body moves fastest at the centre because it has been speeding up the whole way in. So velocity is at its biggest value $v = \omega A$ here.

Step 3: Kinetic energy at the centre.
Since $KE = \tfrac{1}{2}mv^{2}$ and the speed is highest, the kinetic energy is maximum at the mean position.

Step 4: Potential energy at the centre.
$PE = \tfrac{1}{2}kx^{2}$, and at $x=0$ this is zero. So PE is minimum, not maximum.

Step 5: Acceleration at the centre.
$a = -\omega^{2}x$, and at $x=0$ this is also zero. So acceleration is least here, not greatest.

Step 6: Pick the right choice.
Only the statement KE is maximum is true at the mean position. This is option 1.
\[ \boxed{\text{KE is maximum at mean position}} \]