# I Working out distance

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1. Nov 2, 2016

Hello. This is my first post so I apologise if I've put it in the wrong category or done something wrong.

Today in my physics lesson I was given a question where we were supposed to find the depth of a well, and we were given the suvat equations. The question was as follows:

"A stone is dropped down a well. It is heard to hit the bottom after 2.9 seconds. How deep is the well?"

I'm not asking for help with the suvat equations or for the answer it wanted, but I noticed it said "it is heard to hit the bottom," and I was trying to work out how deep the well actually is, considering in reality the speed of the sound travelling back up the well will make a very small, negligible difference, but I've been stumped on how to work this out. This isn't homework/coursework help, I'm just wondering if you could work out the actual depth of the well with this information considering the sound has to get back up too.

Since the highest number of significant figures given here is 2, we'll say that the time it has taken is 2.900 seconds to see the difference, because I'm guessing if it was given to 2sf the depth (also given to 2sf) would be the same.

Thank you

2. Nov 2, 2016

### Staff: Mentor

I'm not quite sure I correctly understood you, esp. your last two lines.

We know the acceleration down the well and the fall takes, say $t_1$ seconds for a depth $d$. Then we know the speed of sound, which travels a time, say $t_2$ along the same distance $d$. Next we know that $t_1+t_2=2.9 s$. That's all which is needed to calculate $d$. You can also calculate a depth $d'$ under the assumption that $t_2=0$. Then $d-d'$ will be the error margin.

3. Nov 2, 2016

I've been taught that if we're only given numbers to 2 significant figures, we can only quote our final answer to 2 significant figures. So if I give the time to more significant figures, the answer will be more precise too. This is what my physics teacher told me, I'm not sure if you'd do this in a real world physics problem.

How would you calculate d? The stone would fall to the bottom of the well, and then as it hits the bottom, the sound will travel back up the well. The total time taken will include this time the sound travels back up. If I was told that the stone hits the bottom at 2.9 seconds later, I'd be easily able to work that out, but I'm not sure how you'd work it out when the 2.9 seconds includes the time taken for it to fall plus the sound travelling back up.

4. Nov 2, 2016

### Staff: Mentor

Normally one doesn't want to know inches in an example like this, so it doesn't matter how long the sound takes. In addition, you would have to take the distance from your hand to your ear into account, the speed of your neural pathways, reaction time, and so on. And as you can see at the list, the most crucial source of error is the measurement itself! This normally dictates how precise the calculation should be.

But as a general advice: Be as accurate as possible as long as possible.
The reason for that is: if you calculate with estimated or rounded figures, the errors throughout the calculation accumulate!

Let's take your example (the easy case):
$d=\frac{1}{2} \cdot g \cdot (2.9\, s)^2$. With a mean value $g=9.80665\, m/s^2$ we get $41.23696325 \,m$. But already this varies from $41.344 \,m$ to $41.12\, m$ depending on where on earth the well is. With $2$ significant figures, i.e. $d=2.9 \cdot 9.8\, m$ we get $41.21\, m$.

So you see that the calculation with two significant figures looks good.
This has two reasons: Firstly, we only have one calculation step, so errors cannot add up. Secondly, we don't really know the actual value of $g$ and the mean value only differs at the fourth significant figure from the rounded one. Both reasons show us, that we were lucky in this case. However, we cannot assume to always be lucky.
What also can be seen, is that already the variation of $g$ due to our lack of knowledge spreads by more than $20 \, cm$. So it makes no sense in this case, to be more exact than, say $d = 41.2\, m \pm 10\, cm$. The advantage of calculating it to $41.23696325 \, m$ is, that we can now round it to our significant figures and get $d=41.2\, m \pm 5\, cm$. Not much of a difference, but rounded upon our decision instead of incorrect estimations during the calculation. Therefore: Calculate as accurate as possible as long as possible. At the end of your calculation, you can adjust the significant figures according to your input level of accuracy. And of course it doesn't make sense to answer: The depth of the well is $41.23696325 \, m$.

In daily life, the error margins depend vastly on the goal one has. Errors in structural calculations of bridges or buildings are potentially fatal, whereas the depth of a well usually is not. In particle physics you want to have as much information on possible errors as you can get.
Constructors of bridges or nuclear power plants calculate with large error margins in order to guarantee safety. Particle physicist calculate as precise as possible. The entire issue of error analysis is a field of research on its own.

5. Nov 2, 2016

6. Nov 2, 2016

### Cutter Ketch

The sound isn't made until the stone hits bottom, so the two transit times are sequential and just add. If the depth of the well is d we have the time to fall is

d = 1/2 g tf2

Solving for tf

tf = sqrt(2 d/g)

If the speed of sound is v the time for the sound to travel up the well is

tu = d / v

The total time is the sum of these two

ttot = d / v + sqrt(2 d / g)

This is a quadratic equation in "sqrt(d)" which can be solved with the quadratic formula. Looking at the result you then try to choose plus or minus by letting v go to infinity and seeing which one recovers the standard form. Now something weird happens. The result is unstable with respect to the speed of sound v as v goes to infinity. The quadratic result goes to 0 over 0. Well, we know what to do with that. We apply L'Hospital's rule and eventually find

d = t2/(2 / g + 4 t / v)

Which now we can see as v goes to infinity returns the familiar form.

7. Nov 5, 2016

What do you mean by "seeing which recovers the standard form"? Is it just which one stays positive?

8. Nov 5, 2016

### Cutter Ketch

I mean we know the answer when the speed of sound is assumed to be infinite and we only calculate the time to reach the bottom of the well. It is d = 1/2 g t2 the standard form of a body under constant acceleration from rest. Knowing that our answer must collapse to this form when we set the speed of sound to infinity is a convenient way (but not the only way) to determine whether to use the plus or minus in the quadratic equation. In this case it had the advantage of revealing why the form from the quadratic equation was confusing. It was because the result was indeterminant at large v. It approached zero over zero.

That doesn't mean it doesn't give the right answer at any finite value of v, but it does suggest there is a better mathematical form for large v which I revealed using L'Hospital's rule. (which I bring up again mostly because I like saying L'Hospital. Such an unlikely pronunciation given the spelling. French is strange)

9. Nov 6, 2016