Q122 / 180 · KCET 2024 · Physics

For a particle executing simple harmonic motion (SHM), at its mean position

  • A
    velocity is zero and acceleration is maximum.
  • Correct answer
    B
    velocity is maximum and acceleration is zero,
  • C
    both velocity and acceleration are maximum.
  • D
    both velocity and acceleration are zero.

Solution

We know that velocity $v$ and acceleration of a particle in SHM. $v=\omega \sqrt{A^2-y^2}$ and $a=-\omega^2 y$ $\therefore$ At mean position, $y=0$ $\therefore \quad v=\omega \sqrt{A^2-0}$ $\Rightarrow \quad v_{\max }=\omega A$ and $a=0$

Hence, the correct answer is option: 'B'.

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