Question

In: Physics

Two sound waves (call them X and Y) travel through the air. Wave X has a...

Two sound waves (call them X and Y) travel through the air. Wave X has a wavelength of 0.1 m and a pressure amplitude of 0.04 Pa. Wave Y has a frequency of 3430 Hz and a pressure amplitude of 4 Pa.

Fill in the following:

The intensity of Y is greater than  that of X by ............... times.

Soundwave Y is ................. dB louder than soundwave X.

Solutions

Expert Solution

Well,
the speed of the sound in air is 343 m/s
by using the wave equation, v = f * lambda
f = v/ lambda
fx = 343 / 0.1 = 3430 Hz
y = 343 / 3430 = 0.1

So intensities are,
X = 0.04/(0.1/3430) = 1372Pa/m^2
Y = 4/(0.1/3430) = 137200Pa/m^2

Therefore,

The intensity of Y is greater than  that of X by 100 times.

Loudness is the characteristic of a sound that is primarily a psychological correlate of physical strength (amplitude). More formally, it is defined as "that attribute of auditory sensation in terms of which sounds can be ordered on a scale extending from quiet to loud".
Loudness, a subjective measure, is often confused with objective measures of sound strength such as sound pressure, sound pressure level (in decibels), sound intensity or sound power. Filters such as A-weighting attempt to adjust sound measurements to correspond to loudness as perceived by the typical human. However, loudness perception is a much more complex process than A-weighting. Loudness is also affected by parameters other than sound pressure, including frequency, bandwidth and duration.


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