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

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Homework Help Overview

The problem involves proving the relationship Vmax = (pi/2)*Vavg for a particle in simple harmonic motion (SHM), where Vavg represents the average speed during one complete cycle of motion. The discussion centers around the definitions and calculations of average velocity and average speed in the context of SHM.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of average velocity and average speed, noting the distinction between the two. One participant attempts to apply the average value function to the velocity equation but encounters confusion regarding the results. Others suggest focusing on the average speed instead of average velocity, prompting further exploration of the definitions involved.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of average speed versus average velocity. Some guidance has been offered regarding the correct definitions, but no consensus has been reached on the approach to proving the relationship.

Contextual Notes

Participants are grappling with the implications of using velocity versus speed in their calculations, which has led to confusion about the average value over a complete cycle. The distinction between average speed and average velocity is a key point of contention in the discussion.

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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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