Q. A mass 1 kg suspended from a spring shoes force constant is 400$\ Nm^{- 1}$, executes simple harmonic oscillation. When the total energy of the oscillator is 2j, the maximum acceleration experienced by the mass will be
A) $2\ {ms}^{- 2}$
B) $4\ {ms}^{- 2}$
C) $40\ {ms}^{- 2}$
D) $400\ {ms}^{- 2}$
Explanation
(c) Energy stored= work done
So $E = \frac{1}{2}kr^{2}$
or $r = \sqrt{\frac{2E}{k}} = \sqrt{\frac{2 \times 2}{400}} = \frac{1}{10}\ m$
$a = \omega^{2}r = \left( \sqrt{\frac{k}{m}} \right)^{2} \times \frac{1}{10}$
$= \left( \frac{400}{1} \right) \times \frac{1}{10} = 40\ ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.