- Home
- Standard 11
- Physics
13.Oscillations
easy
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
$2$
C
$1.5$
D
$1$
Solution
(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$
Standard 11
Physics