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I've tried several times to shout when I'm inside the pool but have failed to make any sound. Nor am I able to hear anyone talking outside. Why does this happen?

The frequency does not change and sound travels faster in water than in air. Partial reflection should take place. So when sound travels from outside, it should reach my ears and I should be make out if someone is talking outside the pool, right?

And when I'm shouting , often water fills up my mouth (which should be due to the pressure of water ) but if it didn't, would my shouts reach outside?

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  • $\begingroup$ Yes, you can make noise underwater by 'shouting'. Higher frequencies work better. Reference - my 8-year old self many years ago. $\endgroup$
    – Jon Custer
    Jan 30, 2016 at 15:15
  • $\begingroup$ Yes but the intensity is quite low and what i'm saying is indistinguishable. $\endgroup$
    – Quark
    Jan 30, 2016 at 15:26
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    $\begingroup$ Well, yes. The concept you are looking for is 'impedance' - the sound velocities in air and water are very different, so they don't couple very well at all. $\endgroup$
    – Jon Custer
    Jan 30, 2016 at 15:28

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That's almost duplicate of this question. You should definitely read it.

It's caused by the change of propagation media. If you shout, the sound is produced by the disturbances of the airflow done by your vocal folds. Again: the airflow. Then it must overcome the huge impedance change to the water (and there comes the loss of power). And from the water to the air surrounding the pool, it will be all over again.

To be illustrative: there is no practical parallel in simple optics. The "refractive index" of the water compared to the air is giant. We thank only our "logarithmic hearing" that we can percipe something.

There is one more case. You can hear yourself shouting under the water very good, but it's caused by sound conduction through your skull bones. See this question for further reference.

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