MCQ
The maximum speed of a particle executing $S.H.M.$ is $1\,m/s$ and its maximum acceleration is $1.57\,m/se{c^2}$. The time period of the particle will be .... $\sec$
- A$1/1.57$
- B$1.57$
- C$2$
- ✓$4$
✓
Answer
Correct option: D.
$4$
d
(d) Given max velocity $\omega a = 1$ and maximum acceleration ${\omega ^2}a = 1.57$
$\therefore \frac{{{\omega ^2}a}}{{\omega a}} = 1.57 $
(d) Given max velocity $\omega a = 1$ and maximum acceleration ${\omega ^2}a = 1.57$
$\therefore \frac{{{\omega ^2}a}}{{\omega a}} = 1.57 $
$\Rightarrow \omega = 1.57$
$ \Rightarrow \frac{{2\pi }}{T} = 1.57$
$\Rightarrow T = 4$
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