CBSE · Class 11 · Physics · OscillationsThe maximum velocity of a particle executing SHM with amplitude $A$ and angular frequency $\omega$ is given by:
Step-by-Step Solution
Velocity in SHM is $v = \omega \sqrt{A^2 - x^2}$. Maximum velocity occurs at the mean position ($x = 0$), giving $v_{\max} = A\omega$.
Detailed Options Breakdown
Option : $A\omega$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 1: $A\omega^2$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Oscillations.
Option 2: $A^2\omega$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Oscillations.
Option 3: $A/\omega$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Oscillations.
💡 Study Guide: This question tests core syllabus concepts from Oscillations. For formulas, key summaries, and mock exam reference guides, read the full Oscillations Revision Notes.