22 Sep 07
1 edit
This one might be the hardest yet...
Waves / Vibrations
13. An object is hung on a spring, and the frequency of oscilation of the system, f, is measured. The object, a second identical object, and the spring are carried to space in the Space Shuttle. The two objects are attached to the ends of the spring, and the system is taken into space on a space walk. The spring is extended, and the system is released to oscillate while floating in space. The coils of the spring do not bump into one another. What is the frequency of oscillation for this system, in terms of f?
22 Sep 07
Originally posted by Ramnedwow, this is much more complicated!
This one might be the hardest yet...
Waves / Vibrations
[b]13. An object is hung on a spring, and the frequency of oscilation of the system, f, is measured. The object, a second identical object, and the spring are carried to space in the Space Shuttle. The two objects are attached to the ends of the spring, and the system is taken into sp ...[text shortened]... nto one another. What is the frequency of oscillation for this system, in terms of f?[/b]
intuition says 1/2 f. the two objects are attached to the same spring. How is it any different than one object on one of the springs?
but, here is where it gets "wierd"
Now you are in space and "weightless" so the idea before, that the object was "falling down" and springing back up is probably irrelevant.
It's probably more like the object was on the spring and well, it oscillated level, like a ball on a table?
Maybe it doesn't even matter at all? maybe an object oscillating on a spring has the SAME frequency regardless of gravity!!!!
so, 1/2 f sounds right, except that assumes equal stretch and contraction. that wouldn't be right, though.
it's way way over my head! 🙄
22 Sep 07
Originally posted by Ramnedw=sqrt(k/m)=2.Pi.f Let's denote 2.Pi=a just to simplify notation. So w=sqrt(k/m)=a.f So f=sqrt(k/m)/a
This one might be the hardest yet...
Waves / Vibrations
[b]13. An object is hung on a spring, and the frequency of oscilation of the system, f, is measured. The object, a second identical object, and the spring are carried to space in the Space Shuttle. The two objects are attached to the ends of the spring, and the system is taken into sp ...[text shortened]... nto one another. What is the frequency of oscillation for this system, in terms of f?[/b]
f'=sqrt(k/m'😉/a=sqrt(k/2m)/a=1/sqrt(2)*sqrt(k/m)/a=f/sqrt(2)
And the symbols all have the usual meanings.
22 Sep 07
1 edit
Originally posted by adam warlockWhat's your answer?
w=sqrt(k/m)=2.Pi.f Let's denote 2.Pi=a just to simplify notation. So w=sqrt(k/m)=a.f So f=sqrt(k/m)/a
f'=sqrt(k/m'😉/a=sqrt(k/2m)/a=1/sqrt(2)*sqrt(k/m)/a=f/sqrt(2)
And the symbols all have the usual meanings.
It's probably better to explain the concepts of that work (Which, I must admit, lost me a bit. B/C f can possibly mean force (F), frequency, or the notation in the problem f... ). 😛
22 Sep 07
Originally posted by RamnedThe answer is f over the square root of two.
What's your answer?
It's probably better to explain the concepts of that work (Which, I must admit, lost me a bit. B/C f can possibly mean force (F), frequency, or the notation in the problem f... ). 😛
And the symbols stand for:
w - angular frequency
sqrt() - square root
k - spring constant
m - mass
. is for multiplication
Pi is for the mathematical constant
f - frequency
Ps: Russ, please give us the math tabs! Pretty please!!! 😵
22 Sep 07
Originally posted by adam warlockSo, are you saying that the system has a frequency higher than f by a factor of r(2)? That's what it looks like.
The answer is f over the square root of two.
And the symbols stand for:
w - angular frequency
sqrt() - square root
k - spring constant
m - mass
. is for multiplication
Pi is for the mathematical constant
f - frequency
Ps: Russ, please give us the math tabs! Pretty please!!! 😵
22 Sep 07
1 edit
Originally posted by RamnedNo. I must be expressing myself wrong. I'm saing that the system has a frequency smaller by a factor of sqrt(2).
So, are you saying that the system has a frequency higher than f by a factor of r(2)? That's what it looks like.
f over square root of 2 doesn't mean f dividing by the square root of 2?
22 Sep 07
1 edit
Originally posted by adam warlockok, ok. I see. So - the clamped system of objects will vibrate with a frequency that is smaller than f by a factor of the sqrt of 2? That's your answer? I need to be sure...if you are correct, then it's going to be necessary to explain your answer in conceptual explanation rather than calculating.
No. I must be expressing myself wrong. I'm saing that the system has a frequency smaller by a factor of sqrt(2).
f over square root of 2 doesn't mean f dividing by the square root of 2?
23 Sep 07
2 edits
Originally posted by adam warlocklol , ok. I'm sorry for my ignorance 😛
Yes.
That is actually incorrect. I don't know where you went wrong, but to hint, you are looking close. When the spring is set into oscillation in space, what's going on with the coil at the exact center? Answering that is perhaps how to begin to logically explain what's going on with the system (And determining the answer). Just a bit of a prompt, since this is a tough one.
23 Sep 07
Originally posted by RamnedOk. I din't thought that one long enough anyway. But I'll give a try now. Just wait a moment. If I go to bed I'll tell you.
lol , ok. I'm sorry for my ignorance 😛
That is actually incorrect. I don't know where you went wrong, but to hint, you are looking close. When the spring is set into oscillation in space, what's going on with the coil at the exact center? Answering that is perhaps how to begin to logically explain what's going on with the system (And determining the answer). Just a bit of a prompt, since this is a tough one.
23 Sep 07
1 edit
Originally posted by adam warlocklol, ok. Maybe someone else can help you. When I was taking my Physics course, I found waves / oscillations to be tough...
I'll have some sleep, it's 1:21 around here, and think about it.
Tomorrow I'll post something. Even if it is just me saying I couldn't crack it.