Question

Question: The intensity at the threshold of hearing for the human ear at a frequency of about \({\bf{1000}}\,{\bf{Hz}}\)is \({{\bf{I}}_{\bf{0}}}{\bf{ = 1}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 12}}}}\,{\bf{W/}}{{\bf{m}}^{\bf{2}}}\), for which\({\bf{\beta }}\), the sound level, is \({\bf{0}}\,{\bf{dB}}\). The threshold of pain at the same frequency is about\({\bf{120}}\,{\bf{dB}}\), or \({\bf{I = 1}}{\bf{.0}}\,{\bf{W/}}{{\bf{m}}^{\bf{2}}}\)corresponding to an increase of intensity by a factor of \({\bf{1}}{{\bf{0}}^{{\bf{12}}}}\)By what factor does the displacement amplitude,\({\bf{A}}\), vary?

Short answer

The displacement amplitude A varies by a factor of\({10^6}\).

Step-by-step solution

Concept

The link between intensity and displacement amplitude is given by the equation:\(I = 2{\pi ^2}v\rho {f^2}{A^2}\)where\(A\)is the displacement amplitude. Thus,\(I \propto A\), or\(A \propto \sqrt I \).

Given Data

The intensity arises by a factor\(I = {10^{12}}\) .

Calculation

The amplitude would increase by a factor of the square root of the intensity increase.

The amplitude increases is,

\(\begin{array}{c}A \propto \sqrt I \\ \propto \sqrt {{{10}^{12}}} \\ \propto {10^6}\end{array}\)

Hence, the amplitude increases by the factor\({10^6}\).