MCQ
The average kinetic energy of a cup doing simple harmonig motion will be :
- ✓$\frac{1}{4} kA ^2$
- B$\frac{1}{2} kA ^2$
- C$kA ^2$
- Dnone of these
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$\text{v}_\text{A}>\text{v}_\text{B}>\text{v}_\text{C}.$
$\text{v}_\text{A}<\text{v}_\text{B}<\text{v}_\text{C}$
$\text{v}_\text{A}=\text{v}_\text{B}=\text{v}_\text{C}$
$\text{v}_\text{B}=\frac{1}{2}(\text{v}_\text{A}+\text{v}_\text{C})$
$\text{2C}\ \text{and}\ \text{2V}$
$\frac{\text{C}}{2}\ \text{and}\ \frac{\text{V}}{2}$
$\text{2C}\ \text{and}\ \frac{\text{V}}{2}$
$\frac{\text{C}}{2}\ \text{and}\ \text{2V}$
