MCQ
A particle executing $S.H.M.$ of amplitude 4 cm and $T = 4 \,sec.$ The time taken by it to move from positive extreme position to half the amplitude is ..... $\sec$
- A$1$
- B$0.33$
- ✓$0.67$
- D$1.22$
$\frac{a}{2} = a\cos \omega t$
$\Rightarrow \cos \omega t = \frac{1}{2}$
$\Rightarrow \omega t = \frac{\pi }{3}$
$ \Rightarrow \frac{{2\pi t}}{T} = \frac{\pi }{3}$
$\Rightarrow t = \frac{{\frac{\pi }{3} \times T}}{{2\pi }} = \frac{4}{{3 \times 2}} = \frac{2}{3}\sec $
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