A particle in SHM has amplitude $A$ and angular frequency $\omega$. Its maximum speed is:
A$A\omega$
B$A\omega^2$
C$A/\omega$
D$\omega/A$
Answer & 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).
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