Challenge Questions

Sounds about like a grandfather clock to me.

Actually, a simple pendulum gives the shortest possible time for a given length, and therefore requires the longest possible length for a given period. In principle, we could make a compound pendulum arbitrarily short. However, the challenge has a drawing that implies a simple pendulum, so your answers (if not all of the text) do appear to correspond to the question as posed**, so you well deserve the G.A. rating.

However, I think you would agree the numerical solution is not very interesting, as it is indistinguishable from the limiting small-amplitude swing (to an accuracy of about a part in 16000).

So what else could the questioner have meant? SlideRuler interpreted the question as "What is the shortest simple pendulum whose swing is limited by a ceiling that could have a period of 2-seconds". Although that does not correspond exactly to the text, I think this more interesting interpretation is likely to be what was intended - given the wording "the shortest" and "based on your answer" - and that it uses of all the features in the drawing.

Regards

Fyz

**Which I translate as "What is the shortest (redundant constraint...) length of an ideal simple pendulum that has a period of 2-seconds when it is released from a 1-degree angle".

I must agree that I have in fact looked for the longest possible length. My reasoning must have been that 1) the pendulum was a simple point mass on the end of a massless string and 2) the period was approximately 2 seconds. This implies only one possible length, being both the longest and shortest and everything in between.

Even if the pendulum were a body with distributed mass, I think the period for the larger angle would be the same as for a simple pendulum.

However, if the pendulum were of distributed mass, I'm also inclined to believe the minimum length could be less. I'm going to have to get out some paper and see if I'm up to the derivation.

Thanks,

Jim

You can see the important features using just two point masses at different distances from the pivot. However, the simple pendulum being the maximum period for the length also depends on the definition of "length".
The usual definition is the distance from the pivot to the centre of gravity.

However, if you use the end-to-end length, the simple pendulum has the longest period - until you place some of the mass above the pivot (which seems to be disallowed by the drawing).

Jim's correct, but because I assume the angle is quite small, I just used the approximations.

Since you state the Length and Time, Acceleration calculates to 9.8696, so you must be on a slightly denser world than here. Maybe that one is 'ideal'.

(Interestingly, if you simply divide 9.8 by 9.8696, you get Jim's 'minimum length'.)

However, given the assumption of staying on the same 'ideal world', my answers are:

1) the period of the swing would still be 2 seconds,

2) the minimum length is still 1 meter

[now sitting anxiously in front of the computer until 29JUL08 ]

The answer to (1) is the same period. The period of a pendulum is the same regardless of the angle thru which is passes.

The answer to (2) is: I don't know.

I believe that Jim35848 is correct in post 1 and I gave him a GA. I thought I would try a different formulation for the result.

I have assumed that the pendulum is made up of string with negligible mass and a point mass at the end so that the centre of mass is at the point.

If we plotted the position of the point on the circle whose radius is the length of the string vs time while it is swinging we should see a sinusoid (sine wave) whose formula would be given by (distance along circle) = Amplitude.sine(Bt). Where B = ((2.pi)/T) radians, t = time and T being the period of the sinewave.

The Amplitude of the sinewave is the maximum distance that it travels along the circle on either side of the midpoint of its swing which equals ((L.2.pi.A)/360) where A = initial angle (in degrees) at which pendulum is released and L = length of the string.

The formula of the sinewave is therefore y = ((L.2.pi.A)/360).sine(Bt) where y = distance along circle. Differentiating wrt t gives the velocity of the point as it swings

= B.((L.2.pi.A)/360).cos(Bt) which has a maximum value when cos(Bt) = 1.

Max velocity V = B.((L.2.pi.A)/360) ….. 1. which occurs at the midpoint of the swing.

Now the gravitational potential energy of the point when it is released is converted to kinetic energy when it is in the midpoint of its swing as this is lowest point. The gravitational potential energy = the difference in height of the point from when it is released to its lowest point (midpoint) = L.(1-cosA).

The gravitational potential energy = m.g.L.(1-cosA) equals the kinetic energy of the point at the midpoint of its swing = ½.m.V². where m = mass of point, g = acceleration due to gravity & V is velocity of point at midpoint of swing.

Therefore V = (2.g.L.(1-cosA))^1/2….. 2.

Equating 1 & 2 and using B = ((2.pi)/T) gives T=(4.pi².L^1/2.A)/(360.(2.g.(1-cosA))^1/2) …..3.

It can be seen from the above that the period of the swing is fairly constant for small values of A but does increase as A increases. The value of T₃/T₁ (where T₃ = period when A=3 degrees and T₁ = period when A = 1 degree) = (3.(1-cos(1))^1/2).(1-cos(3))^-1/2 = 1.0001 which corresponds to a period of 2.0002 seconds when A=3 degrees if the period when A=1 degree was 2 seconds.

Now the value of L can be obtained by substituting T = 2 and A = (1 degree) into formula 3. to get 0.993 metres.

After consultation with Phys I realised my mistake with this formulation. I assumed that the motion of the pendulum could be described by a sinusoid. For this to be true the restoring force (due to component of gravity tangent to the arc of the pendulum) would need to be linearly proportional to the displacement which it clearly isn't. It is approximately true for small angles with an increasing error as the angle increases. It is still better than the simple pendulum approximation.

Hi

I used the aproximation of T = 2*pi*sqrt((L/g)*cos(angle/2))

We assume g = 9.8

For T=2 we have angle=1 degree. From this we can determine L the length.

We get L = 0.99309 m

Then we use L=0.99309 and angle = 3 degrees to get the period = 1.9998 sec

Almost the same periods and this is because of the smallness of the angles.

I would check that formula if I were you. Pendula get slower as the angle increases.

Yes, you are right. Should read T = 2*pi*sqrt(L/(g*cos(angle/2))

and L becomes 0.9929 and the second period becomes 2.0003 sec

Hi FYZ,

I understand that you sent a message to Chris in relation to the answers to some challenge questions. I will check the answer to the Submarine question and I will follow with you.

In relation to the previous question about the electrons in a wire, I created the question will the sole purpose of showing that even if electrons in free "space' can move at a speed close to the speed of light, in a conductors, when an electric field is applied to them, the drift very slow toward the positive terminal of the battery. The idea was not to have the readers to make calculations. I did not give an important parameter needed for the calculation, which is the diameter of the wire. Without this it would be impossible to calculate the time of displacement. I gave the type of wire and its length, just as a diversion tactic.

In my answer I assumed a speed of 0.1 mm/sec just as figure that will help me to show that the time an electron takes to travel a any distance (10 m in our case) is long.

So, this question is not calculational enterprise but only a question that would drive the user to (only) think.

Thanks.

Abe

Hi Abe

Regarding the electron speed:

I didn't feel that those who thought in terms of free-space velocities would be convinced without some basis for the low speed in the conductor. The only respondent (clearly with some sort of experience in the area) who replied along your lines assumed a velocity of 5-mm/second.

But the relevant sums are easy, and such rules of thumb can be highly misleading. The situations in solids where the motion of individual electrons become relevant (rather than the overall current flow) are often when dimensions are in the sub-micron region - and we then expect velocities of 10⁴ to >10⁵ metres/second; although this is well-sub-relativistic, it is rather short of "net velocities in solids being low".

The full formula for the effect of a finite starting angle is given by a Legendre elliptic function of the first kind:
T(φ)/T(0) = 2/∏.∫₀^∏/2(1-sin²(φ/2).sin²(θ))^-1/2dθ

Comparing the answers:

Jim's truncated series is easily within a part in 10¹⁰ of the exact value at 3⁰. The error would rise to 2% at 90⁰.
The error in BobD's answer is -0.006% at 3⁰.
I was not familiar with the approximation given by carel. But it is within a part in 10⁸ at 3⁰ - and may be closer (I've only checked using rather crude tools). At 30⁰ the error is about 0.05%, but (remarkably) it is still within 0.1% at 90⁰. So I've learned something - and I'm giving him a G.A.

I decided to programme this. And I found that I'd slipped up on my transferring of the errors in carel's formula when using the old tool
The errors should be:
.............. carel ............ BobD ........ Truncated
At 3⁰...... 0.7*10^-8 ..... -0.006% ... -3*10^-11
At 30⁰.... 0.9*10^-6 ..... -0.6% ....... -3*10^-5
At 60⁰.... 0.14% ......... -2.6% ...... -0.19%
At 90⁰.... 0.9% ........... -6.7% ...... -2.0%
N.B. The percentages are with respect to the reference period (i.e. the small swing limit)

Not quite as good as I'd thought, but still pretty good - and (unlike a truncated series) both BobD's and carel's approximations go to infinity at 180⁰

Having read your posting (Good Answer, by the way), I noticed two things:
First that the challenge is for the shortest pendulum - following jim35848's selection of a simple pendulum* I had jumped to the longest. (Magicians would doubtless have a field day with me).
Second, the diagram shows the pendulum consisting of a thin line with a blob on the end, suspended beneath a ceiling. So I think you have correctly interpreted the challenge as being a simple pendulum with a 90⁰ limit. Using more terms from the series solution** as a back-up to your RK4, I have a length multiplier of 0.71777, which I imagine is the same as your value (as you didn't state the value used for g)

*As against a compound pendulum, which could of course be as short as you choose
**The zero term is 1, and the nth term is (sinⁿ(θ_max/2).(2.n)!/(2ⁿ.n!)²)²

Thanks for your comments.

I used g = 9.80665m/s

I notice you talk about a length multiplier. whereas I have given an absolute length. Since .718 (your multiplier)*.993(1° amp length)=.713(my answer), I am sure our values are identical.