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

Q: 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

d $\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2} ; \mathrm{U}=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}=\frac{1}{2} \mathrm{m}^{2} \mathrm{x}^{2}$ $\begin{aligned} \therefore \quad \frac{\mathrm{k}}{\mathrm{U}}= \frac{\mathrm{v}^{2}}{\omega^{2} \mathrm{x}^{2}}=\left(\frac{\cos (\mathrm{wt})}{\sin (\mathrm{wt})}\right)^{2} \\ =\cot ^{2}\left(\frac{\pi}{90} \times 210\right) \\= \cot ^{2}\left(2 \pi+\frac{\pi}{3}\right) \\=\left(\frac{1}{\sqrt{3}}\right)^{2}=\frac{1}{3} \end{aligned}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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$2$
D$0.33$

JEE MAIN 2019, Medium

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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