MCQ
A particle undergoing simple harmonic motion has time dependent displacement given by $x(t)\, = \,A\,\sin \,\frac{{\pi t}}{{90}}$. The ratio of kinetic to potential energy $o$ the particle at $t=210\,s$ will be
- A$1/9$
- B$1$
- C$2$
- ✓$0.33$
✓
Answer
Correct option: D.
$0.33$
d
$\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2} ; \mathrm{U}=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}=\frac{1}{2} \mathrm{m}^{2} \mathrm{x}^{2}$
$\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2} ; \mathrm{U}=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}=\frac{1}{2} \mathrm{m}^{2} \mathrm{x}^{2}$
$\begin{aligned} \therefore \quad \frac{\mathrm{k}}{\mathrm{U}}= \frac{\mathrm{v}^{2}}{\omega^{2} \mathrm{x}^{2}}=\left(\frac{\cos (\mathrm{wt})}{\sin (\mathrm{wt})}\right)^{2} \\ =\cot ^{2}\left(\frac{\pi}{90} \times 210\right) \\= \cot ^{2}\left(2 \pi+\frac{\pi}{3}\right) \\=\left(\frac{1}{\sqrt{3}}\right)^{2}=\frac{1}{3} \end{aligned}$
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