MCQ
An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15\, cm/sec$ and the period is $628$ milli-seconds. The amplitude of the motion in centimeters is
- A$3$
- B$2$
- ✓$1.5$
- D$1$
✓
Answer
Correct option: C.
$1.5$
c
(c) ${v_{\max }} = a\omega = a\frac{{2\pi }}{T}$
$ \Rightarrow a = \frac{{{v_{\max }}T}}{{2\pi }} = \frac{{15 \times 628 \times {{10}^{ - 3}}}}{{2 \times 3.14}} = 1.5\,cm$
(c) ${v_{\max }} = a\omega = a\frac{{2\pi }}{T}$
$ \Rightarrow a = \frac{{{v_{\max }}T}}{{2\pi }} = \frac{{15 \times 628 \times {{10}^{ - 3}}}}{{2 \times 3.14}} = 1.5\,cm$
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