Question

A linear harmonic oscillator of force constant $2\times {10}^6\text{N/m}$ and amplitude $0.01\text{m}$ has a total mechanical energy of $160$ Joules. Which of the following is true?

• A. $\text{Maximum potential energy is}100\text{J}$
• B. $\text{Maximum K.E. is}160\text{J}$
• C. $\text{Maximum P.E. is}160\text{J}$
• D. $\text{Minimum P.E. is zero}$

Answer 1

The correct option is C $\text{Maximum P.E. is}160\text{J}$

Harmonic oscillator has some initial elastic potential energy ${\text{PE}}_0$ and amplitude of harmonic variation of energy is
$\frac{1}{2}k{A}^2=\frac{1}{2}\times 2\times {10}^6\times (0.01{)}^2=100\text{J}$
This is the maximum kinetic energy of the ocsillator. Thus, ${\text{KE}}_{max}=100\text{J}$.
But total mechanical energy is $160\text{J}$
As total mechanical energy is conserved
$\therefore {\text{PE}}_0+{\text{KE}}_{max}=160\text{J}$ (at mean position)
$\Rightarrow {\text{PE}}_0={\text{PE}}_{min}=60\text{J}$
We know during oscillation, at some instant entire $\text{KE}$ is coverted to $\text{PE}$
$\therefore {\text{PE}}_{max}={\text{PE}}_0+{\text{KE}}_{max}$
${\text{PE}}_{max}=60\text{J}+100\text{J}=160\text{J}$ (At extreme position)