Question

A linear harmonic oscillator of force constant $2\times {10}^{6}N/m$ and amplitude 0.01 $m$ has a total mechanical energy 160 $J$. Its

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

Answer 1

The correct option is B Maximum K.E. is 100 $J$

$K{E}_{max}=\frac{1}{2}m{v^2}_{max}=\frac{1}{2}m{A}^2{\omega }^2=\frac{1}{2}k{A}^2=100J$
This is the maximum kinetic energy of the oscillator.