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

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SUMMARY

The discussion centers on proving the equation Vmax = (π/2) * Vavg for a particle in simple harmonic motion (SHM). The mathematical model used is x(t) = A*cos(ωt) and the velocity function is v(t) = -Vmax*sin(ωt). The average speed during one cycle is calculated using the average value function, leading to the conclusion that Vavg = (2*Vmax)/π, thus confirming the relationship between Vmax and Vavg.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM) principles
  • Familiarity with calculus, specifically integration and derivatives
  • Knowledge of average value functions in mathematics
  • Ability to manipulate trigonometric functions and their properties
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  • Study the derivation of average speed in simple harmonic motion
  • Learn about the relationship between angular frequency (ω) and maximum velocity (Vmax)
  • Explore the implications of average velocity versus average speed in oscillatory motion
  • Investigate the use of integral calculus in physics problems involving motion
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to simple harmonic motion and average speed calculations.

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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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