MCQ
A particle executes $SHM $ in a line $4\, cm $ long. Its velocity when passing through the centre of line is $12\, cm/s$. The period will be ..... $s$
- A$2.047$
- ✓$1.047$
- C$3.047 $
- D$0.047$
✓
Answer
Correct option: B.
$1.047$
b
(b) Length of the line = Distance between extreme positions of oscillation $= 4 \,cm$
So, Amplitude $a = 2\,cm.$
also ${v_{max}} = 12\,cm/s.$
$\because {v_{\max }} = \omega a = \frac{{2\,\pi }}{T}a$
$ \Rightarrow T = \frac{{2\,\pi a}}{{{v_{max}}}}$ $ = \frac{{2 \times 3.14 \times 2}}{{12}}$$ = 1.047\,\sec $
(b) Length of the line = Distance between extreme positions of oscillation $= 4 \,cm$
So, Amplitude $a = 2\,cm.$
also ${v_{max}} = 12\,cm/s.$
$\because {v_{\max }} = \omega a = \frac{{2\,\pi }}{T}a$
$ \Rightarrow T = \frac{{2\,\pi a}}{{{v_{max}}}}$ $ = \frac{{2 \times 3.14 \times 2}}{{12}}$$ = 1.047\,\sec $
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