91Ó°ÊÓ

The maximum transverse speed of a particle in simple harmonic motion is found in terms of the angular speed $\mathrm{Ó\neg }$and the amplitude $A$as follows:

$v_{\max }={\mathrm{Ó\neg }}^{2}A$$(*)$

We are asked to show that $v_{\max }=(\mathrm{Ï€}/2)v_{\mathrm{avg}}$where $v_{avg}$is the average speed during one cycle of the motion.

The average speed during one cycle of the motion is equal to the distance moved by the particle divided by the time interval required to complete the motion:

$v_{\mathrm{avg}}=\frac{\mathrm{Δ}x}{\mathrm{Δ}t}$

In a complete cycle (time interval$T$), the particle moves through a distance $4A$So

$v_{\text{avg}}=\frac{4A}{T}$

The time period $t$of the motion is related to the angular speed $\mathrm{Ó\neg }$by $T=2\mathrm{Ï€}/\mathrm{Ó\neg }$So

$v_{\mathrm{avg}}=\frac{4A}{2\mathrm{Ï€}/\mathrm{Ó\neg }}$

$=\frac{2\mathrm{Ó\neg }A}{\mathrm{Ï€}}$

$=\frac{2}{\mathrm{Ï€}}\left(\mathrm{Ó\neg }A\right)$

where the term in parenthesis is the maximum speed $v_{\max }$of the particle according to Equation $(*)$Therefore

$v_{\mathrm{avg}}=\frac{2}{\mathrm{Ï€}}{v}_{\max }$

Rearrange for$v_{max}$$v_{\max }=\frac{\mathrm{Ï€}}{2}{v}_{\mathrm{avg}}$