dependence of fundamental frequency of vibration of a stretched string on the medium in which it is kept [closed]

suppose a stretched wire's fundamental frequency in air is 280 Hz. What would be it's fundamental frequency in water ? (all other conditions of the string remain same)

I looked into the laws of vibrations of stretched strings, but all of them give information on characteristics of string, but nothing about the surrounding medium. Please help.

The answer to the question is 243.2 Hz, but I am unable to calculate it myself. I read all the texts of fundamental modes and harmonics but found no way forward.

closed as off-topic by JMac, Jon Custer, Bill N, Michael Seifert, David Z♦Jan 25 '18 at 19:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• One would need assume that the water doesn't affect the damping (pun intended) of the string, which might be included in "all other conditions ...." – Bill N Jan 25 '18 at 18:07
• I'd suggest reading the homework and exercise policy. This doesn't seem like it is on topic here. – JMac Jan 25 '18 at 18:21
• Replacing deleted comment because I think it's important to point out that this question doesn't have a unique answer. I won't put it here a third time if it gets deleted again. If you increase the wire linear mass density and tension without bound, keeping the fundamental frequency and wire diameter the same, the water will cease to have any damping effect at all (frequency in water will be same as in air). If you assume a very light string and low tension, the frequency will decrease by more than claimed. – Ben51 Jan 25 '18 at 19:20
• +1 Voting to reopen. I think this question is conceptual and interesting. The answer is not obvious and not easily found. – sammy gerbil Jan 25 '18 at 22:34
• I am stumped!!! – niels nielsen Jan 26 '18 at 0:13

the surrounding medium has a characteristic acoustic impedance which can be calculated. if that characteristic impedance is close to that of the vibrating string, then two things will occur: first, the string will be strongly damped and second, the mass of the surrounding medium will begin to couple to the mass of the string and the string will act as if its vibrating mass is increased relative to its tension. both of these effects will reduce the natural frequency of the string.

Per Sammy Gerbil's suggestion, I will enlarge upon my answer in this edit:

When a resonant system is coupled to a load that extracts power from it, the width of its resonant response peak is broadened and the location of that frequency peak shifts down to a slightly lower frequency. Immersion in water will extract power from the vibrating string and dissipate it by a variety of mechanisms and therefore its resonant frequency will certainly be reduced upon immersion. The first thing I would try is plugging the string's characteristics into the resonance equation and add progressive amounts of damping, to see how strong the damping effect is and whether or not it can account for the frequency shift.

I'm going off-line now to search for resources and will edit again if I find clues to share.

• I think the asker would like more detailed information about how he can calculate the change in frequency. – sammy gerbil Jan 25 '18 at 22:37
• now that is a nontrivial problem. do you think the impedance-match formalism would be helpful? if it were me, I'd just do the experiment... ;-) – niels nielsen Jan 25 '18 at 22:41
• the fact that the right answer was given to four-place accuracy argues that there is a closed-form equation for that, but I have no idea how it would look. Very cool question. – niels nielsen Jan 25 '18 at 22:43
• I don't know how this is solved. I think the asker is looking for a simple explanation, nothing too technical. I suspect that the person who set the question had a simple but erroneous solution in mind. – sammy gerbil Jan 25 '18 at 22:52
• I can take some guesses but they wouldn't be "answers". first I would assume a small amount of damping and solve for the damped natural frequency, ignoring the mass-coupling effect. perhaps I will edit my answer... – niels nielsen Jan 25 '18 at 23:08