MCQ
The potential energy of a simple harmonic oscillator at mean position is $2\,joules$. If its mean $K.E.$ is $4\,joules$, its total energy will be .... $J$
- A$7$
- B$8$
- ✓$10$
- D$11$
✓
Answer
Correct option: C.
$10$
c
Total energy $=\frac{1}{2} \mathrm{KA}^{2}+\mathrm{U}_{0}$
Total energy $=\frac{1}{2} \mathrm{KA}^{2}+\mathrm{U}_{0}$
$\mathrm{U}_{0}=2 \mathrm{J}(\text { given })$ and $\frac{1}{4} \mathrm{KA}^{2}=4 \mathrm{J}$
So total energy $=8+2=10 \mathrm{J}$
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