MCQ
At time $t = 0$, a simple harmonic oscillator is at its extreme position. If it covers half of the amplitude distance in $1\, second$, then the time period of oscillation is ..... $s$
- A$2$
- B$4$
- ✓$6$
- D$12$
✓
Answer
Correct option: C.
$6$
c
$\cos \alpha=\frac{A / 2}{A}=\frac{1}{2}$
$\cos \alpha=\frac{A / 2}{A}=\frac{1}{2}$
$\alpha=60^{\circ}$
$60^{\circ}$ angle travelled in $1\, sec.$
$360^{\circ}$ angle travelled in $1\, sec.$ $\frac{360}{60} \times 1=6$ sec.
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