String vibration calculation under water

If a string, like a guitar string, vibrates at 5khz in air, what would the frequency be under water? How to calculate?

thanks

Tony

Reply to
TonyJeffs
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5 kHz, but with a lot of damping.

Lance

*****

TonyJeffs thought carefully and wrote on 8/8/2004 11:28 AM:

Reply to
Lance

Not so. The water also provides added mass. For low modes where the axial distance between vibration nodes is at least three diameters, strip theory is a reasonable way to approximate the added mass. In the Blevins book ("Formulas for Natural Frequency and Mode Shape," Van Nostrand 1979), the strip theory added mass per unit length for a circular cross-section is rho*pi*a^2, where rho is the density of water, and 'a' is the radius of the cross-section. Since vibration frequency varies inversely with the square root of mass, the ratio of submerged to in-vacuo frequency is therefore approximately sqrt[(string mass)/(string mass + water mass)]. For example, for a solid steel string in water, where the specific gravity of steel is

7.85, the ratio of submerged to in-vacuo frequency is approximately sqrt(7.85/8.85)=0.94. That is, the frequency is lowered about 6% by the water. The relative effect of the water would be greater if the string were not solid.

Gordon Everstine

Reply to
Gordo

So basically you're saying that some of the water vibrates along with the string? Is this analagous to the way that water flows through a pipe? Like, the water top of bottom of the velocity profile (where water touches the pipe) have zero velocity. Water sticks to the vibrating string the same way?

Hm. Weird.

-Stephen

Reply to
stephenMF

If an object moves in a fluid, the fluid applies a nonuniform pressure to the object. The ratio of pressure to velocity is referred to as the impedance of the fluid. In general, that pressure has a component in-phase with velocity (a damping-like force) and a component 90 degrees out of phase with velocity (a mass-like force). At low frequency, the impedance is mostly mass-like. The net effect is referred to as an "added mass," so it is reasonable to think of a certain amount of fluid moving with the object. The question is: How much fluid? In general, one can compute the added mass by solving a 3-D potential flow problem (Laplace's equation) for the shape in question.

Reply to
Gordo

Thanks Very interesting.

Why then does your voice double(??) in frequency when you inhale helium?

Tony

Reply to
TonyJeffs

I'm definitely no expert, but if I had to guess I would say that it has to do with the speed of sound through the Helium.

Kind of like the doppler effect. Sound waves are created in and are traveling faster through the He, so that when they hit the air the frequency is apparently increased.

Please reply if this is wrong. It's just a guess.

-stephen

Reply to
stephenMF

Regarding my other reply:

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At 25°C, speed of sound in helium is 1015.957m/s which is much faster than that in air (343m/s).

-stephen

Reply to
stephenMF

The jury is still out... URL:

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David A. Smith

Reply to
N:dlzc D:aol T:com (dlzc)

The speed of sound in a medium shouldn't affect its pitch. If you place a tuning fork on a slab of steel and put your ear against the other end, you hear the same pitch through the steel as you do in the air even though the speed of sound is much faster in steel.

Frequency is frequency. The same number of waves are continuously hitting your ear as are being produced by the fork in the same amount of time; you can't lose or gain waves. The waves are just faster. There is no Doppler effect since there is no relative movement between source and listener. A wave may be going faster, but the wave in front of it is going just as fast. The waves will be farther apart (longer wavelength) to compensate for the speed.

speed = frequency x wavelength

The only factors that affect the frequency of a vibrating string are its length, its tension and its mass per unit length. The mass is what we're debating.

Don Kansas City

Reply to
Don A. Gilmore

Stephen, I would've agreed with you last week, but now I dont!

I think the larynx is like a guitar string. The speed of sound in air, water or helium has no effect on it. And the mass of air, water or helium hs little effect.

If it was a resonant tube, like a flute, I think you'd be right.

I think that the seeming change in frequency is to do with the resonance of the body cavities in a different way, though, like the guitar body, and picks out different frequencies other than the fundamental.

But I'm not yet clear on it.

............... I got into this from wondering about various theories of non-neuronal sound filtering in the inner ear; Von Bekesey travelling, and Helmholtz resonator. I'm not convinced by either of them, and don't think Von Bekesey should've got a nobel prize.

Tony I may have to eat my words

Reply to
TonyJeffs

If it was a flute in air, helium or water, for a given frequency, I think the speed of sound would be proportional to the length of the flute.

If its a tensed string, spring, plate, or tuning fork, I now understand from comments here that the speed of sound is not relevant to the frequency.

Tony

Reply to
TonyJeffs

(SNIP)

Right on man. My guess is that it probably has to do with reduced wave pressure on your "voicebox" that allows it to operate at the next mode of vibration (reducing the "effective" mass).

Dave

Reply to
David Harper

It's a bit arrogant of me to disagree with the website, when related processes have just been explained to me, but ;-)

Both ideas on that website are wrong. First, The helium/motor car analogy. It doesn't work. When the larynx generates the sound initially, it travels very fast, and is the equivalent of the stream of cars being verrry far apart. (they overlooked that) When it re-enters air, it slows down and the car spacing, closer together again is back to normal. That's how their analogy would really (not) fit.

Second. On that basis, it's the same as the underwater piano. The speed of sound isn't an issue.

...................... Maybe it's because the voice box acts like a guitar string, rich in harmonics. The trachea acts as a resonant chamber, selectively amplifying harmonics. Changing the characteristics of the resonant chamber, by changing the gas changes the harmonics. If that's the case, then the Mickey mouse voice should be a 3rd, a

5th, or an octave which could easily be checked.

Tony

Reply to
TonyJeffs

Reply to
Anthony Garcia

////

Hmmm.. it depends: Church organ pipes are sensitive to air temperature: How do you think their pitch changes as temperature rises?

Actually, the organ pipe pitch rises with higher temperature, (Even though the pipe's length grows a little....)

....and this is a function of the speed of sound rising with temperature. If you injected an organ pipe with helium, the pitch would indeed rise quite dramatically.

But pipes are standing waves in a gas: and you are focussing on reeds and strings, so the adherant fluid mass is the main ingredient in 'bending' the string frequency, you rightly say.

The human voice in helium has the reed element, and it has the (coupled) tube element too.

Brian W

Reply to
Brian Whatcott

Dear TonyJeffs:

message news:...

Think again. The larnyx is *not* immersed in air (for the example given), and the cavity it resonates against is filled (partially) with helium. So the sound does not "slow back down", but simply slow down. And the resonance of the system is not like a string underwater, but the arrival of pressure waves in a medium with an increased speed of sound, against a flap of skin that is "displaced" by that pressure.

Sounds great!

David A. Smith

Reply to
N:dlzc D:aol T:com (dlzc)

for the musically challenged (me) what guitar string would be able generate the 5kHz?

for the record at this point I'm voting with Lance but with a slight mod.

Wd= Wn {(1- damping^2)}^.5

Reply to
Bob K 207

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