Frequency and Sound Wave Reflection

[Jimmy87]

Homework Statement

Consider two walls that are 128m apart. A student stands in between the two walls such that he is 48m from the left wall. He creates clicks using a loudspeaker and signal generator. He experiences echoes from the two walls. He changes the frequency from 0Hz to a specific frequency f where the two echoes coincide. Calculate the frequency 'f' for when this happens. The speed of sound is 340m/s.

Relevant Equations

speed = distance/time
time period = 1/frequency

I got that the sound wave will take 0.3s between the student and the left wall. It takes 0.5s between the student and the right wall. The first time these waves will coincide is 1.5s (5 trips for left side and 3 for right side). I then did 1/1.5 to get 0.67Hz. However, the answer is 6.67Hz. I'm struggling to visualise exactly what is happening.

Any advice is much appreciated.

 

[haruspex]

Not entirely sure I understand the description. I think it is asking for the lowest frequency at which constructive interference occurs, but that gives a slightly lower answer.
Describing isolated clicks as having a frequency of 0Hz makes no sense to me.

 

[Jimmy87]

Not entirely sure I understand the description. I think it is asking for the lowest frequency at which constructive interference occurs, but that gives a slightly lower answer.
Describing isolated clicks as having a frequency of 0Hz makes no sense to me.

Sorry - I probably didn't remember the question as well as I thought. Here is the actually question below:

 

[haruspex]

There is a crucial difference between that and what you originally posted. It is not just that the echoes coincide with each other; they also coincide with the generation of another click.
Just consider one wall to start with. If each echo coincides with another generated click, what are the possible inter-click times?
Btw, it says the speed of sound to use is 320m/s, and the answer is not 6.67Hz.

 

[Jimmy87]

Thanks. Not quite sure what you mean by possible inter click times? It takes 0.3s to get from the loudspeaker hit the first wall and arrive back and 0.5s for the other wall.

 

[berkeman]

Thanks. Not quite sure what you mean by possible inter click times? It takes 0.3s to get from the loudspeaker hit the first wall and arrive back and 0.5s for the other wall.

And those two times mismatch, no? So what do you have to do so that both sets of echos arrive back at the same time?

 

[Jimmy87]

Wait 5 times for the left wall (1.5s) and 3 times for the right wall (1.5s)?

 

[etotheipi]

Wrong way around!

It suffices to consider the very first emission of sound from the generator at $t=0$, and constrain the times at which the echos return to the generator to be integral multiples of the generator time period $T$ [N.B. this is because you can always shift where you define $t=0$, every emission of sound is equivalent].

Since the time for the first left echo is $t_1 = (3/10) \text{s}$, and the time for the first right echo is $t_2 = (5/10) \text{s}$, you require that for some $m, n \in \mathbb{N}$$$\frac{3}{10} \text{s} = mT, \quad \frac{5}{10} \text{s} = nT$$Can you now figure out what the time period of the generator, $T$, is [i.e. the greatest $T$ that satisfies those equations]?

 

[Jimmy87]

Wrong way around!

It suffices to consider the very first emission of sound from the generator at $t=0$, and constrain the times at which the echos return to the generator to be integral multiples of the generator time period $T$ [N.B. this is because you can always shift where you define $t=0$, every emission of sound is equivalent].

Since the time for the first left echo is $t_1 = (3/10) \text{s}$, and the time for the first right echo is $t_2 = (5/10) \text{s}$, you require that for some $m, n \in \mathbb{N}$$$\frac{3}{10} \text{s} = mT, \quad \frac{5}{10} \text{s} = nT$$Can you now figure out what the time period of the generator, $T$, is [i.e. the greatest $T$ that satisfies those equations]?

The more I think about this the more confused I am getting

For your equation m = 3 and n = 5 so you would get 1/10 in both cases giving 10Hz. However, I have no idea how that equation works and how you know how to set that up. All is I know is still as far as I got originally that to arrive back at the same time you have to wait for 5 echoes off the left wall (5 x 0.3s = 1.5s) and 3 echoes off the right wall (3 x 0.5 = 1.5s). I don't know why you divide each echo time by the opposite integer - only that is satisfies your equation

 

[etotheipi]

The only constraint given in the question is that each echo arriving back at the generator must coincide with a new emission.

If we suppose the time period of the generator is $T$, then the emissions occur at $0, T, 2T, \dots$, etc., i.e. the emissions occur at integral multiples of $T$.

The generator emits its first click at $t=0$. The first echo on the left arrives back at the generator at $t_1 = \frac{3}{10} \text{s}$. In order for this to coincide with an emission, this time $t_1$ must be equal to one of $T, 2T, \dots$ i.e. some integral multiple of $T$. Same goes for the time for the first echo on the right to arrive back at the generator, $t_2$.

Now do you see how we get the equations in #8?

 

[Jimmy87]

The only constraint given in the question is that each echo arriving back at the generator must coincide with a new emission.

If we suppose the time period of the generator is $T$, then the emissions occur at $0, T, 2T, \dots$, etc., i.e. the emissions occur at integral multiples of $T$.

The generator emits its first click at $t=0$. The first echo on the left arrives back at the generator at $t_1 = \frac{3}{10} \text{s}$. In order for this to coincide with an emission, this time $t_1$ must be equal to one of $T, 2T, \dots$ i.e. some integral multiple of $T$. Same goes for the time for the first echo on the right to arrive back at the generator, $t_2$.

Now do you see how we get the equations in #8?

Thanks. I think I'm with you. I kind of now understand your equation in #8 but could you please explain how you figure out m and n? I know they are 3 and 5 respectively but my reasoning for it is wrong as you were saying I had them the wrong way round. So although I have the right answer I think I'm not really sure how?

 

[etotheipi]

Awesome, so we're happy that we can get to $(3/10)\text{s} = mT$ and $(5/10) \text{s} = nT$.

Now notice that the person standing next to the generator is increasing the frequency gradually, which means that they are gradually reducing the time period. When we reach the critical time period, it will be the greatest such time period that satisfies those equations [there will be many values for $T$ that work, but we only want the largest one, i.e. the one we reach first!].

Like you noticed, it's pretty straightforward to see that our $T$ will be $(1/10) \text{s}$. If you want to show this properly, you can just divide the two equations,$$\frac{3}{5} = \frac{m}{n} \implies 3n = 5m$$the lowest integers $m,n \in \mathbb{N}$ that satisfy this will just be $n=5$, $m=3$, and these lowest values of $m$ and $n$ will yield the highest value of $T$, which is what we want. Then you can just plus these back into get $T = (1/10)\text{s}$.

 

[Merlin3189]

The Highest Common Factor (HCF) seems to be what you want. Since these ideas apply to integers, you may have to (pretend to) convert units of course.
So HCF of 0.3 sec and 0.5 sec would have to be HCF of 3 dsec and 5 dsec, or of 300 msec and 500 msec
which come out (since 3 and 5 are prime) as 1 dsec, or 100 msec factoring as 3x100 and 5x100.

If the echo times were less simple, say 0.35 sec and 0.42 sec, then HCF of 35 csec and 42 csec is obviously 7 csec, from 5x7 and 6x7 (If H D d and c are deprecated, then just stick to preferred m or μ or such and get 70 msec)
Even seemingly diffficult times, say 1.05 sec and 1.68 sec yield to simple factorisation.

But of course you need to understand Etotheipi's explanation to see why that works.

(I tried to do a graphical answer, as usual, but realized after I drew the diagram, that I'd used 0.1 sec as the division on my axis and that was why it seemed so easy.)