MCQ
A $S.H.M.$ has amplitude $‘a’$ and time period $T$. The maximum velocity will be
- A$\frac{{4a}}{T}$
- B$\frac{{2a}}{T}$
- C$2\pi \sqrt {\frac{a}{T}} $
- ✓$\frac{{2\pi a}}{T}$
✓
Answer
Correct option: D.
$\frac{{2\pi a}}{T}$
d
(d) ${v_{\max }} = a\omega = \frac{{a\,.\,2\pi }}{T} = \frac{{2\pi a}}{T}$
(d) ${v_{\max }} = a\omega = \frac{{a\,.\,2\pi }}{T} = \frac{{2\pi a}}{T}$
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