Answer
(a) $P_{max} = 7.78~Pa$
This pressure amplitude is less than the pain threshold by a factor of about 0.26.
(b) $P_{max} = 77.8~Pa$
This pressure amplitude is about 2.6 times the pain threshold.
(c) $P_{max} = 778~Pa$
This pressure amplitude is about 26 times the pain threshold.
Work Step by Step
(a) We can find the wave number.
$k = \frac{2\pi~f}{v}$
$k = \frac{(2\pi)(150~Hz)}{344~m/s}$
$k = 2.74~rad/m$
We can find the pressure amplitude.
$P_{max} = ABk$
$P_{max} = (2.00\times 10^{-5}~m)(1.42\times 10^5~Pa)(2.74~rad/m)$
$P_{max} = 7.78~Pa$
We can compare this pressure amplitude to the pain threshold.
$\frac{7.78~Pa}{30~Pa} = 0.26$
This pressure amplitude is less than the pain threshold by a factor of about 0.26.
(b) We can find the wave number.
$k = \frac{2\pi~f}{v}$
$k = \frac{(2\pi)(1500~Hz)}{344~m/s}$
$k = 27.4~rad/m$
We can find the pressure amplitude.
$P_{max} = ABk$
$P_{max} = (2.00\times 10^{-5}~m)(1.42\times 10^5~Pa)(27.4~rad/m)$
$P_{max} = 77.8~Pa$
We can compare this pressure amplitude to the pain threshold.
$\frac{77.8~Pa}{30~Pa} = 2.6$
This pressure amplitude is about 2.6 times the pain threshold.
(c) We can find the wave number.
$k = \frac{2\pi~f}{v}$
$k = \frac{(2\pi)(15,000~Hz)}{344~m/s}$
$k = 274~rad/m$
We can find the pressure amplitude.
$P_{max} = ABk$
$P_{max} = (2.00\times 10^{-5}~m)(1.42\times 10^5~Pa)(274~rad/m)$
$P_{max} = 778~Pa$
We can compare this pressure amplitude to the pain threshold.
$\frac{778~Pa}{30~Pa} = 26$
This pressure amplitude is about 26 times the pain threshold.
