Question

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

A
12
B
14
C
1
D
18

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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