Question

a particle is performing SHM according to equation y=Asin(omega ×t+π/3) . average speed of the particle in time interval t=0 to t=T will be​

Answer 1

Given:

A particle is performing SHM according to equation y=Asin(omega ×t+π/3) .

To find:

Average speed from t = 0 to t = T ?

Calculation:

At t = 0 :

$y1 = A \sin( \omega t + \dfrac{\pi}{3} )$

$\implies \: y1 = A \sin( 0+ \dfrac{\pi}{3} )$

$\implies \: y1 = A \sin( \dfrac{\pi}{3} )$

$\implies \: y 1= \dfrac{ \sqrt{3} A}{2}$

At t = T :

$y 2= A \sin( \omega t + \dfrac{\pi}{3} )$

$\implies \:y 2= A \sin( \omega T + \dfrac{\pi}{3} )$

$\implies \:y 2= A \sin( \dfrac{2\pi}{T} \times T + \dfrac{\pi}{3} )$

$\implies \:y 2= A \sin( 2\pi + \dfrac{\pi}{3} )$

$\implies \:y 2= A \sin(\dfrac{\pi}{3} )$

$\implies \: y2= \dfrac{ \sqrt{3} A}{2}$

So, this since y1 = y2 , this means that the total distance is equal to one full oscillation.

$\therefore \: d = 4 \times (amplitude)$

$\implies \: d = 4A$

Now, average speed:

$v_{avg} = \dfrac{total \: distance}{time}$

$\implies v_{avg} = \dfrac{4A}{T}$

$\implies v_{avg} = \dfrac{4A}{( \frac{2\pi}{ \omega}) }$

$\implies v_{avg} = \dfrac{2A \omega}{\pi}$

So, final answer is:

$\boxed{ \bf v_{avg} = \dfrac{2A \omega}{\pi} }$

Answer 2

$\huge \boxed{ \bf Answer} : \: \: \: \sf \frac{2A \omega}{\pi}$