How many times do a couple of bells ring close enough that they are heard as one?

Question

This is a problem from the book: Recreations in the theory of numbers by Albert Beiler.

The problem is this: You have 2 bells $B_1,B_2$.

$B_1$ rings every $\frac{4}{3}$ seconds while $B_2$ rings every $\frac{7}{4}$seconds.

How many strokes are heard during 15 minutes if 2 strokes following each other within an interval of $\frac{1}{2}$ second or less are perceived as one sound?

I'm not sure how to formalize this problem into a mathematical statement.

Since $\frac{4}{3} = \frac{16}{12}$ and $\frac{7}{4}=\frac{21}{12}$, calculating the least common multiple of $(16,21)=336,$ I arrived at the conclusion that $B_1$ and $B_2$ will ring at exactly the same time after $28$ seconds. In this interval, $B_1$ will have rang $21$ times while $B_2$ $16$ times, so if I find how many times in this interval, both $B_1$ and $B_2$ stroke within $\frac{1}{2}$ second, I could get the total number of strokes heard during the whole period, but I'm not sure how to do this without explicitly finding it out manually.

I did the calculation and found out that the total strokes heard in one 28 second period is $24$, and they meet $12$ times. I get the correct answer but I would like to know if there is a less tedious way of finding the answer

Any help would be appreciated!

Answer 1

Let's make it easier, for the sake of brainstorming, let's rewrite the question by multiplying everything by $12$. One bell rings every $16$ seconds and the second every $21$ seconds and we want to find out how often then ring within $6$ seconds of each other during a $336$ second cycle.

This is more a question of how often they don't.

We need to find times when the $16m - 21n > 6$ and $21(n+1) - 16(m) > 6$. That way you hear the $m$th chime of the $16$ second bell. Or we need to find $21n - 16m > 6$ and $16(m+1) - 21n > 6$. That way you hear the $n$th chime of the $21$ second bell.

The former is saying $16m \equiv 7,....,14\mod 21$. In the $336$ second cycle Each $16m \equiv k \mod 21$ occurs exactly once so this occurs $8$ (out of $21$) times. (The remaining $13$ chimes are within six seconds of the other bell.)

That later is saying $21n \equiv 7,8,9\mod 16$. This occurs $3$ (out of $16$) times. (The remaining $13$ chimes are within six seconds of the other bell.) Thus in the $336$ seconds we hear the bells distinctly $8+3 =11$ times. And of the remaining $16-3 = 13$ chimes of the times the faster bell and the remaining number $21 - 8= 13$ chimes of the slower bell, we hear two chimes as one. So in a $336$ second cycle we hear the bells distinctly $11$ times and as one $13$ times.

So that's our $336$ second cycle.

Now in the $180*60= 10800$ seconds we have $32 \frac {48}{336}$ cycles. So we hear $32(11+13)$ plus however many we hear in $48$ seconds.

In $48$ seconds we hear the $16$ and $21$ bells as one. Then we hear the $32$ second bell by itself and then the $42$ and the $48$ as one.

So we hear $32(11+13) + 3=771$ strokes.

[Rewrite every thing above replacing the word "second" with "twelth second unit".]

Answer 2

In $28''$ bell $B_1$ rings $21$ times, and its sound covers $21''$ of the $28''$. The probability that a random sound of bell $B_2$ (a Dirac $\delta$) is not heard then is ${3\over4}$. During these $28''$ bell $B_2$ emits $16$ Dirac-sounds, only $4$ of which will then actually be heard (elementary number theory takes care of that). It follows that during $28''$ we will actually hear $21+4=25$ strokes; makes $803.57$ in $15'$.

Note that we cannot assume that the bells are "in phase". In fact we have to assume that at the beginning of the $28''$ the phases of $B_1$ and $B_2$ are uniformly distributed in $\bigl[0,{4\over3}\bigr]$, resp. $\bigl[0,{7\over4}\bigr]$. This is taken care of in the above approach.