MCQ
The displacement of an object attached to a spring and executing simple harmonic motion is given by $ x= 2 \times 10^{-9}$ $ cos$ $\;\pi t\left( m \right)$ .The time at which the maximum speed first occurs is
- A$0.25$
- ✓$0.5$
- C$0.75$
- D$0.125$
$\therefore v=\frac{d x}{d t}=2 \times 10^{-2} \pi \sin \pi t$
For the first time, the speed to be maximum,
$\sin \pi t=1$ or, $\sin \pi t=\sin \frac{\pi}{2}$
$\Rightarrow \pi t=\frac{\pi}{2} \quad$ or, $t=\frac{1}{2}=0.5 \mathrm{sec}$
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