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A particle undergoing simple harmonic motion has an amplitude of $10 \mathrm{~cm}$. When the particle is at a displacement of $6 \mathrm{~cm}$ from the centre, then the ratio of its kinetic energy to potential energy is
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The correct answer is:
$16: 9$
Amplitude of particle performing SHM,
$$
A=10 \mathrm{~m}
$$
Instantaneous displacement of the particle,
$$
x=6 \mathrm{~cm}
$$
Kinetic energy of the particle,
$$
K=\frac{1}{2} m \omega^2\left(A^2-x^2\right)
$$
Potential energy of particle,
$$
\begin{aligned}
U & =\frac{1}{2} m \omega^2 x^2 \\
\frac{K}{U} & =\frac{\frac{1}{2} m \omega^2\left(A^2-x^2\right)}{\frac{1}{2} m \omega^2 x^2} \\
& =\frac{A^2-x^2}{x^2}=\frac{A^2}{x^2}-1 \\
& =\left(\frac{A}{x}\right)^2-1=\left(\frac{10}{6}\right)^2-1=\frac{25}{9}-1 \\
\Rightarrow \quad \frac{K}{U} & =\frac{16}{9} \Rightarrow K: U=16: 9
\end{aligned}
$$
$$
A=10 \mathrm{~m}
$$
Instantaneous displacement of the particle,
$$
x=6 \mathrm{~cm}
$$
Kinetic energy of the particle,
$$
K=\frac{1}{2} m \omega^2\left(A^2-x^2\right)
$$
Potential energy of particle,
$$
\begin{aligned}
U & =\frac{1}{2} m \omega^2 x^2 \\
\frac{K}{U} & =\frac{\frac{1}{2} m \omega^2\left(A^2-x^2\right)}{\frac{1}{2} m \omega^2 x^2} \\
& =\frac{A^2-x^2}{x^2}=\frac{A^2}{x^2}-1 \\
& =\left(\frac{A}{x}\right)^2-1=\left(\frac{10}{6}\right)^2-1=\frac{25}{9}-1 \\
\Rightarrow \quad \frac{K}{U} & =\frac{16}{9} \Rightarrow K: U=16: 9
\end{aligned}
$$
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