Vibration velocity

In fact, in physics, acceleration, a = dv/dt and v = dx/dt

Bear in mind of these relations since they are important in the following proof.


To begin with, please refer to the following figure:

http://i726.photobucket.com/albums/ww265/physicsworld2010/physicsworld01Feb181935.jpg

Suppose a particle moves around a circle of radius r from P to Q as shown above. By definition of linear velocity, v = dx/dt

For small θ, x = arc PQ = rθ

Therefore, v = d(rθ)/dt

Since r is a constant, v = r dθ/dt

v = rω

Where ω is angular frequency (or angular velocity), which is defined as ω = 2πf, where f is the frequency.


So, by definition, linear acceleration a is defined as a = dv/dt

Please take a look of the following figure

http://i726.photobucket.com/albums/ww265/physicsworld2010/physicsworld02Feb181935.jpg


Consider a particle moving in a circular path from A to B with a uniform speed v (Velocity is not uniform since the direction is changing). The change in velocity of the particle is:

△v = vsinθ - (-vsinθ) = 2vsinθ

If the particle takes time t to travel from A to B, the acceleration of the particle is

a = 2vsinθ / t

Also, by the relation vt = 2rθ

a = 2vsinθ / (2rθ / v) = v^2/r X sinθ/θ

Taking θ→ 0

the instantaneous acceleration of the particle is

lim θ→ 0 a

= v^2 / r lim θ→ 0 (sinθ/θ)

= v^2 / r (Since lim θ→ 0 (sinθ/θ) = 1)

We have: a = v^2 / r

As by above, v / r = ω

So, we have acceleration, a = ωv = 2πfv

And then we have finished the proof.


P.S., For SHM, it is just a projection of a particle undergoing uniform circular motion, it is the same for SHM.