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The total mechanical energy of a particle executing simple harmonic motion is E. When the displacement is half the amplitude its kinetic energy will be

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$\frac{3 E}{4}$

$\frac{E}{2}$

$\frac{E}{4}$

answer is A.

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Detailed Solution

$E = \frac{1}{2} {mω}^{2} A^{2} and K . E = \frac{1}{2} {mω}^{2} (A^{2} - x^{2})$

Putting $x = \frac{A}{2} or x^{2} = \frac{A^{4}}{4}$

K.E. = $\frac{1}{2} {mω}^{2} (A^{2} - \frac{A^{2}}{4}) = \frac{3}{4} [\frac{1}{2} {mω}^{2} A^{2}] = \frac{3 E}{4}$

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