MCQ
A particle starts oscillating simple harmonically from its equilibrium position then, the ratio of kinetic energy and potential energy of the particle at the time $T/12$ is : ($T =$ time period)
- A$2 : 1$
- ✓$3 : 1$
- C$4 :1$
- D$1 : 4$
✓
Answer
Correct option: B.
$3 : 1$
b
since $x=a$ sin $\omega t,$ we have
since $x=a$ sin $\omega t,$ we have
$x\left(t=\frac{T}{12}\right)=a \sin \left(\frac{2 \pi}{T} \times \frac{T}{12}\right)=a \sin \left(\frac{\pi}{6}\right)=\frac{a}{2}$
then the ratio of the kinetic energy to the potential energy at $x=\frac{a}{2}$ will be given by
$\frac{K . E .}{P . E .}=\frac{\frac{1}{2} k\left(a^{2}-x^{2}\right)}{\frac{1}{2} k x^{2}}=\frac{a^{2}-x^{2}}{x^{2}}$
$\Rightarrow \frac{K . E .}{P . E .}=\frac{a^{2}-\left(\frac{a}{2}\right)^{2}}{\left(\frac{a}{2}\right)^{2}}=\frac{3}{1}$
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