A man stands on the ground at a fixed distance from a siren which emits sound of fixed amplitude . The man hears the sound to be louder on a clear night than on a clear day. Why?
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$\begingroup$ I don't think this is a realistic scenario. In fact I would wager it is usually the other way around. See physics.stackexchange.com/questions/133198/… $\endgroup$– ProfRobCommented Feb 16, 2015 at 15:53
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3 Answers
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The speed of sound depends on the square root of temperature, so the refractive index is proportional to $T^{-1/2}$.
Let's assume that the sound is emitted isotropically. During the day, the usual situation is that the temperature decreases with height. Thus the refractive index increases with height. This will tend to make sound waves emitted in the direction of the listener bend upwards into the atmosphere - reducing the amplitude/loudness that they hear.
At night it is quite possible to get a temperature inversion (especially a clear night), such that air near the ground is colder than higher up.
As the refractive index decreases with height it means that sound waves propagating upwards at some angle to the horizontal will be bent back towards the ground. The sound waves at some distance from the source will be more intense than you might expect if the waves propagated isotropically.
So I think the situation would be completely contrary to what you say in your question - and indeed that has been my empirical experience.
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$\begingroup$ In other words, the acoustic impedance near the ground at night is lower? $\endgroup$– SparklerCommented Feb 16, 2015 at 16:06
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Sound is a longitudinal wave whose amplitude stands for its "loudness". When sounds travels in medium, it attenuates, i.e. becomes less loud, exponentially, with an exponential coefficient, $\alpha$. So what you're actually interested in is the change in $\alpha$ with temperature.
For newtonian fluid for example, this coefficient is equal to:
$$\alpha = \frac{2 \eta\omega^2}{3\rho V^3}$$
And we also know the influence of temperature on the dynamic viscosity of air:
$$\eta \propto \frac{T^{3/2}}{T+120}$$
Also, $\rho\propto T^{-1}$ and the speed is invertially proportional to the square density $V \propto \frac{1}{\sqrt{\rho}} \propto \sqrt T$ so in total
$$\alpha \propto \frac{\eta}{\rho V^3} \propto \frac{\frac{T^{3/2}}{T+120}}{T^{-1}T^{3/2}} = \frac{T}{T+120}$$
So for a neutonian fluid (such as air) the lower the temperature, the lower the attenuation (graph), i.e. at night the siren is louder, in contrast to what your question suggests.
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1$\begingroup$ Agreed, but I think this is a minor contributor compared to the anisotropy introduced by a changing refractive index with height. $\endgroup$– ProfRobCommented Feb 16, 2015 at 15:55
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$\begingroup$ @RobJeffries, I assumed that the distance between the man and the horn is not that large for the altitude effect to "kick in". Nontheless, the difference in amplitudes would be very small in my approach. $\endgroup$– SparklerCommented Feb 16, 2015 at 15:57
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When sound propagates in the lower atmosphere it can travel through denser or thinner parts of air. This could cause a "portion" of the sound to bend upwards, making it harder to hear.
Factors that could cause the bending are: wind speed and/or temperature that change based on altitude.For more info you can check this link.
Additionally, wind sound can partially mask the siren sound, that is, make it appear less loud than it actually is (or even make it completely inaudible).
In the experiment you describe, night and day would have an effect on both temperature and winds, therefore can potentially cause the sound to be refracted upwards, changing the sound intensity reaching the observer.
