JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 2)
In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.
$${1 \over 2}$$
$${3 \over 4}$$
$${1 \over 3}$$
$${1 \over 4}$$
Explanation
$$K = {1 \over 2}m{\omega ^2}({A^2} - {x^2})$$
$$ = {1 \over 2}m{\omega ^2}\left( {{A^2} - {{{A^2}} \over 4}} \right)$$
$$ = {1 \over 2}m{\omega ^2}\left( {{{3{A^2}} \over 4}} \right)$$
$$K = {3 \over 4}\left( {{1 \over 2}m{\omega ^2}{A^2}} \right)$$
$$ = {1 \over 2}m{\omega ^2}\left( {{A^2} - {{{A^2}} \over 4}} \right)$$
$$ = {1 \over 2}m{\omega ^2}\left( {{{3{A^2}} \over 4}} \right)$$
$$K = {3 \over 4}\left( {{1 \over 2}m{\omega ^2}{A^2}} \right)$$
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