Simple Harmonic Motion Average Velocity

[tridianprime]

Homework Statement

At time t = 0, a point starts oscillating on the x - axis according to the law x = a sin(ωt). Find the average velocity vector projection (I assume it means magnitude based on previous questions in the book).

Homework Equations

The Attempt at a Solution

I knew that the average velocity over a large motion will be the same as the average velocity over a 1/4 of an oscillation so I let x = a. Then a = a sin(ωt) ⇒ t = π/2ω and so the average velocity is 2aω/π.

However, in the official solution it claims the answer is ((2√2)/3)*aω. I don't see how this makes sense and fear I am either missing something substantial or the official solution is incorrect.

 

[BvU]

Hi tri,

Is the problem statement complete ? the average velocity over a long period of time would end up at zero, wouldn't it ?

 

[tridianprime]

Hi tri,

Is the problem statement complete ? the average velocity over a long period of time would end up at zero, wouldn't it ?

The problem statement is complete in the sense that the book says no more. I took velocity vector projection and thought the same but I believe, based on previous questions, it is referring to the magnitude.

 

[BvU]

OK, so what is wanted is the mean of the absolute value of the velocity.
As you say, averaging over one quarter period is OK.

⇒ t = π/2ω and so the average velocity is aπ/2ω.

Something goes wrong here. The velocity is ##a\omega\cos(\omega t)##. The average for the cosine gives you a number between 0 and 1.
a and ##\omega## can't end up one as numerator and the other as denominator (on different sides of the dividing line).
For one, it would not yield the dimension of velocity !

 

[tridianprime]

OK, so what is wanted is the mean of the absolute value of the velocity.
As you say, averaging over one quarter period is OK.
Something goes wrong here. The velocity is ##a\omega\cos(\omega t)##. The average for the cosine gives you a number between 0 and 1.
a and ##\omega## can't end up one as numerator and the other as denominator (on different sides of the dividing line).
For one, it would not yield the dimension of velocity !

Sorry, I mistyped - it is now fixed. 2aω/π I also fixed another ambiguity in my typing - I hope it is now all unambiguous.

 

[BvU]

That's what I get too. The ##{2\over 3}\sqrt 2 \approx 0.94 ## is a mystery to me: rather close to 1 !

 

[tridianprime]

I suppose it's probably a book error then (with 1500 problems you expect there to be a few, it is irodov's problem book (for reference)) - I doubt we have both forgotten something important, especially with such a simple problem. Thanks.

 

[Fightfish]

This is the statement of the original problem from Irodov, I think:

It would seem that Irodov meant the "averaged over 3/8 of the period after the start" to apply for all three parts and not just the last part. When you do that, you should arrive at the given solution. Might have been a transcription error when they translated the original Russian text.

 

[BvU]

The problem statement is complete in the sense that the book says no more

 

[tridianprime]

I didn't have that part - I only noted it down from a lesson and was working off of that. Sorry.

@Fightfish Thanks for the clarification

 

[theodoros.mihos]

The symbol is ##\langle v_x\rangle##. May this mean ## \langle v(x)\rangle##.