Last night, while watching a movie on TV, the scary bad guy managed to dive off of a cliff, and catch the good guy by suprise. the good guy was submerged in the water, no more that 20 feet away.
good guy is underwater > bad guy lands in water no more that 20’ away, after diving off of a high cliff, without the good guy noticing ( apparently.)
I made a note of pointing out that there was no way the good guy could have NOT heard the bad guy crash into the water. I said water is a better transmitter of sound that air. My friend disagreed, but didn’t say why. He’s a very smart person, and generally tends to be right. I told him it was true, and that even as a child I got to experience the effect. With an electronic watch sumberged under water on the other side of the pool, it’s alarm sounded. I was submerged on the other side of the pool, and the sound waves seemed LOUDER.
I have been unable to confirm whether or not the sound was louder. What I have found is that sound travels much faster in liquids(1500m/s) than in gases (350m/s).
Is the watch easier to hear under water, or through the air.
Yes. Sound is generally transmitted faster and better the more dense the medium (there are other factors).
As far as the real(!)-world scenario of the movie is concerned, I suppose it could be argued that the guy in the water didn’t hear the other guy dive in against whatever ambient noise was present; are we talking about a cliff overhanging a small, calm pond, or a cliff overhanging a choppy ocean, with waves breaking on the rocks?
It was a cliff overhaning a choppy ocean, but I think that it’s besides the point.
I understand that solids are better that liquids, which is better that gases for transmitting sound. In terms of volume, would the watch in the example be louder through 50 feet of water, or 50 feet of air?
The situation in your OP seems a typical example of movie (or TV) physics - i.e. in no way connected to the real world, and I agree with you that provided the bad guy hit the water at speed the good guy should have heard the splash. However, I’d like to point out that your experience with the watch alarm in the swimming pool is not exactly transferrable here. The swimming pool (with your ears underwater ) acts like an enclosed room, so you will also be hearing reflected sound waves from the sides and bottom, and to a lesser extent from the water-air interface of the surface of the pool.
Mangetout is right that in general the velocity of the sound waves and the efficiency of propagation increases with the density of the medium. So with other factors being as equal as possible - say you drain the swimming pool and repeat your watch alarm experiment, the sound will be louder when the pool is full.
The speed of sound in water is much higher than in air, and I believe attenuation is much lower, so sound travels farther and faster. One interesting consequence of this is that it is very difficult to localize sounds heard underwater. Our brain localizes sounds by comparing the time shift between the signals from the two ears in much the same way it creates 3D vision by comparing the inputs from two eyes. Since the speed of sound is higher underwater, the time differences between the two ears is much smaller, so it is very difficult to localize. Your good guy might have heard a splash but been unable to tell what direction it came from.
It may be beside the point of the question “Is sound more efficiently transmitted through water than air”, but it not beside the point of “Shouldn’t you hear someone jumping into the water 20 ft. away from you.” The ocean is a really noisy place, and waves breaking against rocks is one of the louder things you can hear underwater. You could easily lose the sound of a 170 lb entering the water against the sound of hundreds of tons of water crashing against a solid surface.
The other factor is that it’s very difficult to determine the direction of a sound when underwater. The 2 tricks you use to figure out the direction of a sound - louder in one ear than other + sound arrives at one ear sooner than the other - don’t work too well when submerged.
Don’t you hate it when your friends just always have to be right?
I was not disputing that the guy in the water would not have heard the diver.
I was disputing the blanket statement that sound travels better through water than air. According to this site, that is a misconception.
I still maintain that the main reason the beeping watch seems louder underwater is because there is no ambient noise down there except for the sound of the pool filters (since you have decreased hearing sensitivity underwater), and as moes lotion said, the water/air boundary acoustically encloses the pool.
But I am willing to be proven wrong. I have tried googling for hard numbers on acoustic attenuation in air and underwater to no avail. If someone can dig up a cite or otherwise prove to me that sound travels more efficiently in water than in air, I’ll give Casey $5 bucks (No, the SOFAR channel doesn’t count).
Attenuation in water is a complex phenomenon. A common formula used is that of Schulkin and Marsh. It factors the temperature and salinity of the water as well as the frequency of the sound. However, other factors are at play, notably the pH of the water, for sounds under 1000 Hz. Some reading. This site has a table of attenuation coefficients in water according to that formula.
In air, attenuation is a function of temperature, humidity, barometric pressure and sound frequency. There’s a nifty site, here, that allows you to calculate the attenuation coefficients in air.
The first table linked above gives an attenuation coefficient of 0.0558 dB/kyd (kilo yard) for a sound at 3500Hz and water at a temperature of 70º F.
The calculator in the other site returns an a.c. of 0.023527 dB/m for the same 3500Hz sound in air, temperature = 20°C, humidity = 50%, pressure = 101 kPa.
Now, the fact that attenuation in water is measured in decibels per thousands of yards, while in air it’s decibels per metres should tell you something. Water is hugely more conductive to sound waves than air, even if attenuation varies quite a bit in both mediums.