MCQ
In an angular $SHM$ angular amplitude of oscillation is $\pi $ $rad$ and time period is $0.4\,sec$ then calculate its angular velocity at angular displacement $ \pi/2 \,rad$. ..... $rad/sec$
- A$34.3$
- ✓$42.7$
- C$22.3$
- D$50.3$
✓
Answer
Correct option: B.
$42.7$
b
$\omega=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{0.4}=5 \pi$
$\omega=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{0.4}=5 \pi$
$\Omega=\omega \sqrt{\theta_{0}^{2}-\theta^{2}}$
$\Omega=5 \pi \sqrt{\pi^{2}-\frac{\pi^{2}}{4}} \Omega=5 \pi \sqrt{\pi^{2}-\frac{\pi^{2}}{4}}$
$=\frac{5 \pi^{2} \sqrt{3}}{2}=42.7 \mathrm{rad} / \mathrm{s}$
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