MCQ
$y = 2\, (cm)\, sin\,\left[ {\frac{{\pi t}}{2} + \phi } \right]$ what is the maximum acceleration of the particle doing the $S.H.M.$
- A$\frac{\pi }{2}\,cm/s^2$
- ✓$\frac{\pi^2 }{2}\,cm/s^2$
- C$\frac{\pi^2 }{4}\,cm/s^2$
- D$\frac{\pi }{4}\,cm/s^2$
✓
Answer
Correct option: B.
$\frac{\pi^2 }{2}\,cm/s^2$
b
$y = 2\,\sin \,\left( {\frac{{\pi t}}{2} + \phi } \right)$
$y = 2\,\sin \,\left( {\frac{{\pi t}}{2} + \phi } \right)$
velocity of particle $\frac{{dy}}{{dt}} = 2 \times \frac{\pi }{2}\,\cos \,\left( {\frac{{\pi t}}{2} + \phi } \right)$
acceleration $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}}=-\frac{\pi^{2}}{2} \sin \left(\frac{\pi \mathrm{t}}{2}+\phi\right)$
Thus $a_{\max }=\frac{\pi^{2}}{2}$
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