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

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In summary, the homework statement is trying to find the average velocity of a particle in simple harmonic motion. The problem is that there is no SHM with a Vmax of 0.
  • #1
MrMaterial
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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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  • #2
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.
 
  • #3
As Fightfish says it is the average speed.
The instantaneous speed is VmaxSinθ, you can average the speed from 0 to ∏.
 
Last edited:
  • #4
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!
 
  • #5


Hello,

Thank you for your question. First, let's clarify some things. Vmax is the maximum velocity of the particle in SHM, while Vavg is the average speed during one cycle of the motion. These two are not the same thing. Velocity is a vector quantity, while speed is a scalar quantity. Therefore, we cannot simply compare Vmax and Vavg directly. Instead, we need to look at their magnitudes.

To prove the given statement, we will use the fact that the average value of a sinusoidal function over one period is equal to its amplitude divided by π/2. This is a well-known mathematical property that can be proved using calculus. So, let's start with the SHM equation:

x(t) = A*cos(ωt)

We can differentiate this to get the velocity function:

v(t) = -A*ω*sin(ωt)

Now, let's calculate the average speed over one cycle:

Vavg = 1/(2π)*∫|v(t)|dt

= 1/(2π)*∫|A*ω*sin(ωt)|dt

= A*ω/(2π)*∫sin(ωt)dt (since |sin(x)| = sin(x) for all real values of x)

= A*ω/(2π)*(-cos(ωt)) (using the integration formula for sin(x))

= A*ω/(2π)*(-cos(2π) + cos(0)) (evaluating the integral from t=0 to t=2π)

= A*ω/(2π)*(-1 + 1) (since cos(2π) = cos(0) = 1)

= 0

Therefore, the average speed during one cycle of SHM is 0. This may seem counterintuitive, but remember that speed is a scalar quantity and it does not take into account the direction of motion. In SHM, the particle moves back and forth between two points, so its average displacement over one cycle is 0. This means that the average speed over one cycle is also 0.

Now, let's look at the maximum velocity, Vmax. This occurs at the extreme points of the motion, where the velocity is either maximum or minimum. At these points, the velocity is equal to the amplitude multiplied by the angular frequency (A*ω). So, we can write:

Vmax = A*ω
 

1. What is Vmax and Vavg in the context of SHM?

Vmax (maximum velocity) and Vavg (average velocity) are two important parameters in the study of Simple Harmonic Motion (SHM). Vmax refers to the maximum velocity reached by a particle in SHM, while Vavg refers to the average velocity over one complete cycle of motion.

2. How is the equation Vmax = (pi/2)*Vavg derived?

The equation Vmax = (pi/2)*Vavg is derived from the mathematical expression for velocity in SHM, which is given by V = Aωcos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase angle. By taking the average of this equation over one complete cycle of motion (from t = 0 to t = T), we get Vavg = (2Aω)/π. And by substituting ω = 2πf and T = 1/f, we get Vavg = (2πA)/T. Finally, by setting Vavg = Vmax and T = 1/f = 2π/ω, we get Vmax = (pi/2)*Vavg.

3. Can you provide an example of how this equation is used?

Let's say we have a simple pendulum with an amplitude of 10 cm and a frequency of 1 Hz. Using the equation Vmax = (pi/2)*Vavg, we can calculate that the maximum velocity of the pendulum is (pi/2)*(2π*10)/1 = 31.4 cm/s. This means that at the lowest point of the pendulum's swing, the velocity will be 31.4 cm/s. On the other hand, the average velocity over one complete swing will be (2π*10)/1 = 62.8 cm/s. This equation helps us understand the relationship between Vmax and Vavg in SHM.

4. What are the units of Vmax and Vavg?

Both Vmax and Vavg have units of velocity, which are typically measured in meters per second (m/s) or centimeters per second (cm/s), depending on the system of units being used.

5. Are there any exceptions to this equation?

Yes, there are a few exceptions to the equation Vmax = (pi/2)*Vavg. One exception is in the case of damped harmonic motion, where energy is lost due to friction or other external forces. In this case, Vmax will be less than (pi/2)*Vavg. Another exception is in the case of forced oscillations, where an external force is applied to the system, causing Vmax to be greater than (pi/2)*Vavg.

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