First slide

Simple hormonic motion

Question

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

Easy

A

$π^{2} {ma}^{2} v^{2}$
B

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

$4 π^{2} {ma}^{2} v^{2}$
D

${2π}^{2} m a^{2} v^{2}$

Solution

$x = A \cos ωt \Rightarrow v = \frac{dx}{dt} = - Aω \sin ωt KE = \frac{1}{2} {mv}^{2} = \frac{1}{2} {mA}^{2} ω^{2} \sin^{2} ωt Average KE = \frac{1}{2} {mA}^{2} ω^{2} < \sin^{2} ωt > = \frac{1}{2} {mA}^{2} ω^{2} \times \frac{1}{2} = π^{2} {ma}^{2} v^{2}$

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