# Prove that Vmax = (pi/2)*Vavg (SHM)

1. Jun 11, 2012

### MrMaterial

1. The problem statement, all variables and given/known data
For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion.

2. Relevant equations

x(t) = A*cos(ωt) (SHM mathematical model)
v(t) = -Vmax*sin(ωt)

Fave = 1/(b-a)∫f(x)dx

3. The attempt at a solution

As soon as i get started with this problem, I hit a brick wall.

I don't know if this is due to me being brain-dead because of all the studying I've been doing today, but whatever it is I can't seem to wrap my head around it!

the problem: How do I calculate the average velocity?

I know v(t) is the derivative of x(t)

I also know the average value function shown above.

1.) find derivative of x(t) to get v(t)

v(t) = -Vmax*sin(ωt)

2.) use average value function on v(t) to find "average velocity"

since Vavg is defined as the average velocity of one cycle, and one cycle = 2pi

b = 2pi

a = 0

Fave = 1/(2pi - 0)*∫v(t)dt = 0

this makes the equation impossible to prove! no SHM has a Vmax of 0!

yet..... how could the average velocity NOT be zero?

What am I doing wrong here?

2. Jun 11, 2012

### Fightfish

Notice the question says "Average Speed" .
The average velocity is zero simply because for the 2nd half of the period, the oscillator is travelling in the opposite direction! So velocity is 'positive' for one half and 'negative' for the other half, and not surprisingly the average over 1 cycle in zero.

3. Jun 11, 2012

### azizlwl

As Fightfish says it is the average speed.
The instantaneous speed is VmaxSinθ, you can average the speed from 0 to ∏.

Last edited: Jun 11, 2012
4. Jun 11, 2012

### MrMaterial

gah curse them for using v to denote a speed quantity!

my head still doesn't want to work properly, but i understand taking the value of Vmax*sin(θ), although I want to write it as Vmax*sin(ωt), from pi to 0.

I guess ωt = radians = θ

And then for some reason i wanted to use the average value function except with b = pi rather than 2pi. Something tells me that would bring me a wrong answer

Let's see.

Vmax = 20m/s

b = pi
a = 0

(∫20*sin(θ))/pi = 12.73m/s

Guess not!

so Vavg = |(Acos(pi)-Acos(0))/pi|

using the initial formula Vavg = (2*Vmax)/pi

pit them together:

(2*Vmax)/pi = (Acos(pi)-Acos(0))/pi

pi's cross out

2*Vmax = Acos(pi)-Acos(0)

plugging in numbers, everything seems to work out!