What is the definition of a way and how does it relate to travel?

[mancity]

Homework Statement

Determine the period of an 80 cm long pendulum in free fall near Earth's surface.

Relevant Equations

T=2pi*sqrt(L/g)

I put 0, but that is incorrect. Why is 0 an incorrect answer? This is confusing, as if the pendulum is in free fall, wouldn't there be no SHM at all?

 

[Tom.G]

What does your Relevant Equation evaluate to when you put the numbers in it?

 

[kuruman]

Zero period means that if you displace it from the equilibrium position, it takes no time at all to return back. Does this make sense for a pendulum in free fall?

 

[Orodruin]

Zero period means that if you displace it from the equilibrium position, it takes no time at all to return back. Does this make sense for a pendulum in free fall?

Yes! I agree with the OP, it is a very poorly worded question.

If the pendulum is in free fall, then $g = 0$ in its rest frame (which is accelerating relative to the Earth).

A pendulum which is supported at its fixed point is not in free fall. It is acted upon by a supporting force at its fixed point and therefore gravity is not the only force acting upon it.

 

[PeroK]

Homework Statement: Determine the period of an 80 cm long pendulum in free fall near Earth's surface.
Relevant Equations: T=2pi*sqrt(L/g)

I put 0, but that is incorrect. Why is 0 an incorrect answer?

I don't think that 0 is a valid period.

This is confusing, as if the pendulum is in free fall, wouldn't there be no SHM at all?

That's what I think - the equivalence principle applies. And, if it's not SHM, then it makes no sense to specify a period.

 

[256bits]

Homework Statement: Determine the period of an 80 cm long pendulum in free fall near Earth's surface.
Relevant Equations: T=2pi*sqrt(L/g)

I put 0, but that is incorrect. Why is 0 an incorrect answer? This is confusing, as if the pendulum is in free fall, wouldn't there be no SHM at all?

What course are you taking?
And under what section of the course is this question applied.
Certainly something to do with gravity.

The one thing I can say is that if the tension in the string from support to the bob is 0, which could be if the string broke, then the bob would be in free fall.

 

[Orodruin]

The one thing I can say is that if the tension in the string from support to the bob is 0, which could be if the string broke, then the bob would be in free fall.

Or if the entire pendulum is in free fall, as specified by the problem

 

[kuruman]

I put 0, but that is incorrect. Why is 0 an incorrect answer? This is confusing, as if the pendulum is in free fall, wouldn't there be no SHM at all?

OP put in zero for the answer and that is indeed an incorrect answer for the period when $g=0.$ For $g=0$, one gets "infinity" for the period $T=2\pi\sqrt{L/g}.$ This agrees with observation. In free fall, or in free space, if one ties one end of a string to a bob, the other to a support (whatever that means) and gently displaces the bob, the bob will stay where it is. In other words it will take infinite time to return to its initial position.

To answer OP's first question, "why is 0 an incorrect answer?", zero would be the correct answer for the period if the length of the pendulum were zero. Then, if displaced from equilibrium, gravity or no gravity, the bob will take zero time to be returned to the equilibrium position because it's already there!

To answer OP's second question, as explained above, there is no SHM at the two limiting values of the period, zero and infinity. However, there will be SHM for any value of $g$ other than zero or infinity.

Yes! I agree with the OP, it is a very poorly worded question.

The wording is fine. If the question is poorly worded, how could it be made better? If one omitted the bone of contention, "in free fall", the question would become a plug-and-chug and totally boring problem. In its current form it is more interesting. Look at all the discussion it has already triggered.

 

[Orodruin]

The wording is fine. If it's poorly worded, how could it be made better? If one omitted the bone of contention, "in free fall", the question would become a plug-and-chug and totally boring problem. In its current form it is more interesting. Look at all the discussion it has already triggered.

I disagree. The ”in free fall” is misleading. There are two options: Either ut actually means ”in free fall” in which case there is no SHM or what is meant is is the plug and chug with the intention of the pendulum actually being supported in which case it is clearly not in actual free fall.

That it has generated discussion is not a sign of being an interesting problem as much as it being an ill-defined problem in this case - thus generating discussion regarding what the problem actually asks (never a good sign!)

 

[Hill]

In any case, zero is a wrong answer because period is defined as a non-zero number.

 

[PeroK]

In any case, zero is a wrong answer because period is defined as a non-zero number.

Hence, somewhat surprisingly, a constant function is not a periodic function.

 

[kuruman]

I disagree. The ”in free fall” is misleading. There are two options: Either ut actually means ”in free fall” in which case there is no SHM or what is meant is is the plug and chug with the intention of the pendulum actually being supported in which case it is clearly not in actual free fall.

My point is that, taken at face value, the statement of the problem says and means that the system is in free fall and therefore there is no SHM (first option). I also argue and explain that the correct answer is "The period is infinite", not zero as OP put in.

Obviously we will not know the intentions of the author until we hear from OP what the supposedly correct answer is, if OP knows. I am only guessing that the author of the problem intended to write a question that would make one think beyond the plug-and-chug situation which has very little pedagogical value.

 

[Hill]

Hence, somewhat surprisingly, a constant function is not a periodic function.

I thought it is periodic with any non-zero number being a period. (?)

 

[PeroK]

I thought it is periodic with any non-zero number being a period. (?)

Sorry, yes, I meant it doesn't have a fundamental period.

Ironically, its set of periods is $(0, \infty)$, which excludes the only two answers being put forward here!

 

[Orodruin]

I am only guessing that the author of the problem intended to write a question that would make one think beyond the plug-and-chug situation which has very little pedagogical value.

Unfortunately, my posterior tells me that the number of instructors writing interesting questions is far outweighed by the number of instructors writing ill-defined questions. I’d be rather incluned to believe that any interesting discussion has arisen despite the author’s wording rather than thanks to it.

Technically a pendulum in free fall could have any period (the motion being circular with constant angular velocity). A constant solution of course also exists, but that can also be considered to have any period as them $\theta(t+T) = \theta(t)$ for all $T$, including arbitrarily small ones.

 

[jbriggs444]

If the "pendulum" were further stipulated to be a uniform thin rod with a small initial deflection from vertical, it should be straightforward to calculate its period under tidal gravitation.

 

[Orodruin]

Zero period means that if you displace it from the equilibrium position, it takes no time at all to return back.

Also, regarding this, a function $f(t)$ having period $T$ means returning to the original position in time $T$, ie, $f(t+T) = f(t)$. The equilibrium position does not necessarily need to be involved. In the one-dimensional case yes, but consider something like a spherical pendulum, where you can have periodic motion without ever passing the equilibrium point (or motion in a Kepler potential where you can have periodic motion without an equilibrium point at all).

 

[kuruman]

Also, regarding this, a function $f(t)$ having period $T$ means returning to the original position in time $T$, ie, $f(t+T) = f(t)$. The equilibrium position does not necessarily need to be involved. In the one-dimensional case yes, but consider something like a spherical pendulum, where you can have periodic motion without ever passing the equilibrium point (or motion in a Kepler potential where you can have periodic motion without an equilibrium point at all).

And what does zero period mean in these cases? It means that the spherical pendulum is at the point of support (or that the orbiting mass at the force center) and stays there.

My statement

Zero period means that if you displace it from the equilibrium position, it takes no time at all to return back.

was intended to convince the OP that the period cannot be zero when $g=0$, but I can see how it can be misconstrued.

 

[Herman Trivilino]

Homework Statement: Determine the period of an 80 cm long pendulum in free fall near Earth's surface.
Relevant Equations: T=2pi*sqrt(L/g)

I put 0, but that is incorrect. Why is 0 an incorrect answer?

Because $g=0$ and the question asks for the value of $T$.

This is confusing, as if the pendulum is in free fall, wouldn't there be no SHM at all?

Correct.

 

[Orodruin]

This is confusing, as if the pendulum is in free fall, wouldn't there be no SHM at all?

I mean, if we are supposed to get technical … the pendulum is only in approximate SHM for small amplitudes even when it is not in free fall …

 

[PeroK]

And what does zero period mean in these cases?

If zero is allowed, then all motion has a period of zero.

 

[Herman Trivilino]

Ironically, its set of periods is $(0, \infty)$, which excludes the only two answers being put forward here!

True, but it is common for introductory physics instructors and students to ignore such technicalities. For example, the OP states that $T=2\pi\sqrt{\frac{L}{g}}$, something that is seen in virtually every introductory physics textbook at both the high school and college level, but the correct relation is $T\approx2\pi\sqrt{\frac{L}{g}}$, the smaller the amplitude the better the approximation, but there is no value of the amplitude for which $T=2\pi\sqrt{\frac{L}{g}}$, strictly speaking, especially in view of the fact that the period can never equal zero.

This is often the case that a simple question made for beginning students is over-thought by experts. Don't get me wrong, sometimes these types of analyses can lead to improvements in the wording of the question, so in that sense they are useful. But also it is often the case that the in-depth analysis simply confuses the beginner student.

I'm not advocating that these types of analyses be eliminated, I'm just saying that if we're trying to help a beginner, they often don't serve that purpose. I'm sure everyone will agree that for an oscillating simple pendulum $T\rightarrow \infty$ as $g \rightarrow 0$.

 

[PeroK]

I'm sure everyone will agree that for an oscillating simple pendulum $T\rightarrow \infty$ as $g \rightarrow 0$.

A limit, however, does not always describe the limiting case. When $g = 0$, we don't have SHM. And any talk of an oscillation with an infinite period is meaningless.

 

[mancity]

What course are you taking?
And under what section of the course is this question applied.
Certainly something to do with gravity.

The one thing I can say is that if the tension in the string from support to the bob is 0, which could be if the string broke, then the bob would be in free fall.

I am taking AP Physics 1, and currently on Unit 7 (Simple Harmonic Motion)
I have attached the problem statement and the possible answer selections below. "undefined" never appeared as one of the answer choices, so I picked 0 as that was the only other answer that 'made sense' to me.

 

[kuruman]

I am taking AP Physics 1, and currently on Unit 7 (Simple Harmonic Motion)
I have attached the problem statement and the possible answer selections below. "undefined" never appeared as one of the answer choices, so I picked 0 as that was the only other answer that 'made sense' to me.

And, considering it was in freefall, wouldn't there be no SHM at all? Which would effectively make the period of the SHM none, too?
View attachment 337731

It looks like a bad question to me. The third answer from the top would be the correct answer if the pendulum were not said to be "in free fall". Maybe the author has a different understanding of "free fall" from everybody else.

 

[berkeman]

Well, if you delete "in free fall" from that question, one of the other 3 options is correct. Can you ask your instructor what "in free fall" means in this problem statement? Like "letting go" of the pendulum at a small initial angle maybe?

 

[berkeman]

Maybe the author has a different understanding of "free fall" from everybody else.

I thought maybe it could be a language translation issue, but nope, when I look up the OP's location using my Mentor superpowers, they are about 50 miles from me here in Silicon Valley...

 

[PeroK]

Don't try this on the ISS!

 

[kuruman]

they are about 50 miles from me here in Silicon Valley...

I don't know if that's good or bad, but I wouldn't get closer to them than that.

 

[nasu]

Well, if you delete "in free fall" from that question, one of the other 3 options is correct. Can you ask your instructor what "in free fall" means in this problem statement? Like "letting go" of the pendulum at a small initial angle maybe?

This will also make sense of the "near Earth's surface". If it is really in free fall it is a completely irrelevant addition to the statement.

 

[Orodruin]

I rest my case!

it is a very poorly worded question.

 

[Herman Trivilino]

The third answer from the top would be the correct answer if the pendulum were not said to be "in free fall". Maybe the author has a different understanding of "free fall" from everybody else.

More likely, there was some kind of editorial error made when the question was written. Perhaps a prior version contained that phrase and it was determined that there was no correct answer. So they decided to remove it and it never got removed in a latter re-write.

I bet that the grading algorithm would mark it correct if a student chose the third answer.

 

[kuruman]

I entered in the Google search engine the statement of the problem in (quotation marks) and got two hits. One was our PF thread and the other from coursehero.com. This is a site where students get help with their course work for a fee.

I did not find the question because they wanted me to sign up and the display was blurred until I did. This is a site where students get help with their course work for a fee. However, I found an interesting tool that I tried for free. It is an AI powered paraphraser! Just what is needed to further students' education. I tried it (free trial offer) by entering the last paragraph in my post #12. Here are the results. Draw your own conclusions.

My text from post #12
Obviously we will not know the intentions of the author until we hear from OP what the supposedly correct answer is, if OP knows. I am only guessing that the author of the problem intended to write a question that would make one think beyond the plug-and-chug situation which has very little pedagogical value.

AI Paraphraser, a.k.a. Quillbot, text
https://www.coursehero.com/tools/paraphraser/
Naturally, until we find out from OP—assuming OP is aware of the ostensibly correct response—we will not be aware of the author's intentions. I can only speculate that the problem's author wanted to pose a query that would force readers to consider options other than the plug-and-chug scenario, which is incredibly low value in terms of education.

 

[256bits]

Or if the entire pendulum is in free fall, as specified by the problem

Theoretically, microgravity on the order of 1^-5
would produce a tension in the string while the ensemble is in free fall.

Practically, difficult to achieve a measurement due to
- difficulties with modelling the structure as being rigid or elastic.
- Initial conditions of support of the ensemble either being suspended from the top or resting on a base.

 

[Herman Trivilino]

I entered in the Google search engine the statement of the problem in (quotation marks) and got two hits.

About 10 years ago I taught a SAT physics test prep short course for pre-meds. The textbook we used had questions with errors like this peppered throughout. It was a terrible eye-opening experience. Despite that I did it two more times because the pay was outrageously high. Grant money.

 

[mancity]

The correct solution, as given by the system, is 1.8s.

 

[kuruman]

I heard back from my instructor today, and the correct solution, as given by my instructor, is to directly apply the formula for the period of a pendulum to get 1.8s.

That's what I was afraid of, and thank you for asking. I don't mean to be pushy but, if you get a chance, do you mind asking your instructor what kind of information is conveyed by the "free fall" addition to the statement of the problem? Just in case you see "free fall" again in the future, mind you. I think all of us would like to know what your instructor thinks and says about it.

 

[nasu]

This is not necessarily the correct solution but the solution which will give you the credit for that question. He can decide about the marking scheme but this decision will not make a faulty question right.

 

[hutchphd]

Is the course titled "Koan Physics"? Seems to me a badly misguided attempt by someone........

 

[haruspex]

The correct solution, as given by the system, is 1.8s.

Ah, so "free fall" just means nothing is getting in the way of the pendulum.

@laser: that was intended as humour

 

[Orodruin]

Ah, so "free fall" just means nothing is getting in the way of the pendulum.

Ftfy

 

[kuruman]

Ah, so "free fall" just means nothing is getting in the way of the pendulum.

No strings attached?

 

[haruspex]

No strings attached?

"way: a road, track, or path for travelling along."