Question

For a particle performing linear S.H.M. show that its average speed over one oscillation is
$\frac{2aw}{\pi}$
where a is amplitude of S.H.M.


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Answer 1

Concept:

• SHM is the case of periodic motion, in which particle periodically move to and fro around a fixed point know as mean position.
• Net displacement of the particle from mean position (starting point) to mean position (ending point) is Zero, then the average velocity is also zero.
• But, net distance is not zero.

★ Net distance =

• mean to 1st extreme + 1st extreme to mean + mean to 2nd extreme + 2nd extreme to mean
• A + A + A + A = 4A

Solution:

$\implies\rm average\: speed = \dfrac{net\: distance}{time\: taken}$

$\to \rm V_{av} = \dfrac{4A}{T}$

•• Total time taken in one oscillation is known as Time period = T = $\rm \dfrac{2\pi}{\omega}$

$\to \rm V_{av} = \dfrac{4A}{\dfrac{2\pi}{\omega}}$

$\to \rm V_{av} = \dfrac{4A\omega}{2\pi}$

$\to \bf V_{av} = \dfrac{2A\omega}{\pi}$

Hence proved!