​ For a particle performing linear SHM, show that its average speed over one oscillation is 2ωA/π ,where A is the amplitude of SHM. ​

Question

For a particle performing linear SHM, show that its average speed over one oscillation is \(\frac{2\omega A}\pi\),where A is the amplitude of SHM. 

OR 

Show that the average speed of a particle performing SHM in one oscillation is \(\frac2 \pi\) × maximum speed.

Answer 1

During one oscillation, a particle performing SHM covers a total distance equal to 4A, where A is the amplitude of SHM. The time taken to cover this distance is the period (T) of SHM.

Average speed = \(\frac{distance\,covered\,in\,one\,oscillation}{time\,taken\,for\,one\,oscillation}\)

\(\therefore\)v_av = \(\frac{4A}T\)

But T = \(\frac{2\pi}{\omega}\)

Where \(\omega\) is a constant related to the system.

 \(\therefore\)v_av = = 4A x \(\frac{\omega}{2\pi}\) = \(\frac{2\omega A}{\pi}\)

But \(\omega\)A = maximum speed

\(\therefore\) Average speed = \(\frac{2}{\pi}\) x maximum speed