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Question

When the displacement is half of the amplitude, the ratio of potential energy to the total energy is


A
12
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B
14
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C
1
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D
18
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Solution

The correct option is B $$\dfrac {1}{4}$$
Potential energy is given by  $$U_{P} = \dfrac {1}{2} m\omega^{2} y^{2}$$
where, $$y$$ is the displacement of particle from its mean position.
Here $$y = \dfrac {A}{2},$$
$$ \therefore U_{P} = \dfrac {1}{2}m\omega^{2} \dfrac {A^{2}}{4}$$
Total energy is given by  $$E = \dfrac {1}{2}m\omega^{2} A^{2}$$
$$ \Rightarrow \dfrac {U_{P}}{E} = \dfrac {1}{4}$$.

Physics

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