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

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The discussion revolves around proving that Vmax equals (pi/2) times Vavg for a particle in simple harmonic motion (SHM). The participant initially struggles with calculating average velocity, noting that the average velocity over one cycle is zero due to the oscillation's direction change. They clarify that the question refers to average speed, not average velocity, which is a crucial distinction. By integrating the velocity function and applying the average value concept correctly, they find that Vavg can be expressed in terms of Vmax, leading to the desired relationship. Ultimately, the calculations confirm that Vmax is indeed (pi/2) times Vavg, resolving the initial confusion.
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Homework Statement


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.


Homework Equations



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


Fave = 1/(b-a)∫f(x)dx

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? :confused:
 
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Notice the question says "Average Speed" .
The average velocity is zero simply because for the 2nd half of the period, the oscillator is traveling 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.
 
As Fightfish says it is the average speed.
The instantaneous speed is VmaxSinθ, you can average the speed from 0 to ∏.
 
Last edited:
gah curse them for using v to denote a speed quantity! :mad:

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!:approve:

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!
 
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