A particle of mass m executes SHM with amplitude 'a' and frequency 'v'. The average kinetic energy during motion from the position of equilibrium to the end is:

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$2 π^{2} {ma}^{2} v^{2}$

$\frac{1}{4} {ma}^{2} v^{2}$

$π^{2} {ma}^{2} v^{2}$

$4 π^{2} {ma}^{2} v^{2}$

answer is B.

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Detailed Solution

$\overset{Average}{⟶} K \cdot E = \frac{K \cdot E_{max} + 0}{2}$
$\begin{array}{l} Average K.E = \frac{1}{2} K \cdot E_{max} = \frac{1}{2} \times \frac{1}{2} m ω^{2} A^{2} \\ = \frac{1}{4} m 4 π^{2} v^{2} a^{2} = π^{2} {ma}^{2} \cdot v^{2} \end{array}$

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