Q. A spring executes SHM with mass of $10kg$ attached to it. The force constant of spring is $10N/m.$ If at any instant its velocity is 410cm/s, the displacement will be (where amplitude is $0.5m$)
A) $0.09\ m$
B) 0.3 m
C) 0.03 m
D) 0.9 m
Explanation
(b) Angular velocity$\ \omega = \sqrt{\left( \frac{k}{m} \right)} = \sqrt{\left( \frac{10}{10} \right)} = 1$
Now $v = \omega\sqrt{a^{2} - y^{2}} \Rightarrow y^{2} = a^{2} - \frac{v^{2}}{\omega^{2}} = (0.5)^{2} - \frac{(0.4)^{2}}{1^{2}}$
$\Rightarrow y^{2} = 0.9 = y = 0.3\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.