Question

A linear harmonic oscillator of force constant $2\times {10}^{6}N{m}^{-1}$ and amplitude 0.01 m has a total mechanical energy 160 J. which of the following statements are correct?
(i) Maximum PE is 100 J (ii) Maximum KE is 100 J
(iii) Maximum PE is 160 J (iv) Minimum PE is zero

• A. Both (i) and (iv)
• B. Both (ii) and (iii)
• C. Both (i) and (ii)
• D. Both (ii) and (iv)

Answer 1

The correct option is B Both (ii) and (iii)

Total mechanical energy is $E_T=160J$
$\therefore U_{max}=160J$
At extreme position KE is zero. Work done by spring force from extreme position to mean position is $W=\frac{1}{2}k{A}^2$
$\therefore K_{max}=W=\frac{1}{2}(2\times {10}^6)(0.01{)}^2=100J$
$\therefore U_{min}=160-100=60J$