Physics · Energy in S.H.M.: kinetic and potential energies
Potential energy of a simple harmonic oscillator at its mean position is 0.4 J.
Potential energy of a simple harmonic oscillator at its mean position is 0.4 J. If its kinetic energy at a displacement half of its amplitude from mean position is \( 0.6 \mathrm{J}, \) its total energy is
- A. \( 1.0 J \)
- B. \( 1.2 J \)
- C. \( 1.4 J \)
- D. 1.6 .5
Step-by-step solution
Let total energy E = KE + U. At mean position (x=0), U = 0.4 J (constant offset). At amplitude (x=A), U_max = (1/2)kA^2 + 0.4 J, and since KE=0, E = U_max. At x=A/2, U = (1/2)k(A/2)^2 + 0.4 = (E-0.4)/4 + 0.4 = E/4 + 0.3. Given KE = 0.6 J at x=A/2, so E = 0.6 + E/4 + 0.3 → (3/4)E = 0.9 → E = 1.2 J.
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