Question

A mass of $1kg$ suspended from a spring whose force constant is $400{Nm}^{-1}$, executes simple hormonic oscillation. When the total energy of the oscillation is $2J$, the maximum acceleration experienced by the mass will be:

• A. $2$ ${ms}^{-2}$
• B. $4$ ${ms}^{-2}$
• C. $40$ ${ms}^{-2}$
• D. $400$ ${ms}^{-2}$

Answer 1

The correct option is C $40$ ${ms}^{-2}$

Let the amplitude of the oscillation be $A$

Energy of oscillation is $2J$

$\therefore$ $E=\frac{1}{2}k{A}^2$ where $k=400N{m}^{-1}$

$2=\frac{1}{2}\times 400\times A^2$ $⟹A^2=0.01$ OR $A=0.1$ $m$

Also $k=m{\omega }^2$

$\therefore$ $400=1\times {\omega }^2$ $⟹{\omega }^2=400$

Maximum acceleration $a={\omega }^{2}A=400\times 0.1=40$ $m{s}^{-2}$