alleryStart learning →A particle in SHM has amplitude $A$ and angular frequency $\omega$. Its maximum speed is:
- $A\omega$
- $A\omega^2$
- $A/\omega$
- $\omega/A$
Show answer and solution
Correct answer: A. $A\omega$
Position $x(t) = A \sin(\omega t + \phi)$. Velocity is its time derivative: $v(t) = A\omega \cos(\omega t + \phi)$. Maximum value is $v_{\max} = A\omega$, achieved when the particle passes through the mean position ($x = 0$).
Maximum acceleration is at the extremes: $a_{\max} = A\omega^2$ (option B is the acceleration max, the common trap).