Question

For each part, intersperse your equations with written explanation as to why your approach makes sense in terms of physical concepts. Treat each part as a “concept question”

3) A mother bat is looking for her baby by echolocation. The baby bat chirps at fb = 20 kHz. A distance of d = 1 m from the baby, its chirp has an intensity of βd = 60 dB relative to a standard intensity of I0 = 1 pW/m2 (the threshold of hearing). Take the speed of sound to be vs = 343 m/s.

(i) The mother hears her baby’s chirp at f = 19 kHz. Is the baby flying away or towards her mother? At what speed? Express your answer in terms of f, f0, and vs, and then substitute numbers. You should find 18 m/s.

(ii) What is the total noise power emitted by the baby bat? Give your answer in terms of d, βd, and I0, and then substitute numbers. You should find 12.5 µW.

(iii) The mother hears the baby’s chirp at β = 30 dB (again relative to I0). How far away is the baby? Express your answer in terms of d, β, and βD, and then substitute numbers. You should find 31.6 m.

Answer 1

An approaching source moves closer during period of the sound wave so the effective wavelength is shortened, giving a higher pitch since the velocity of the wave is unchanged. Similarly the pitch of a receding sound source will be lowered.

(a) The baby bat chirps at f_b = 20 kHz. the speed of sound to be v_s = 343 m/s. The mother hears her baby’s chirp at f = 19 kHz which is lower than the baby bat. Therefore, the baby bat is flying away from the mother bat.

$f=\frac{v_{s}}{v_{s}+v}f_{0},\ or\ \frac{v_{s}+v}{v_{s}}=\frac{f_{0}}{f}$

$\therefore \frac{343+v}{343}=\frac{20}{19}$

$\therefore v=\frac{343}{19}=18.05\ m/s$

(b) Sound power is related to sound intensity: P = AI, A is the area and I is the intensity. The reference intensity I₀ = 1 pW/m² . The intensity level is defined as

$\beta _{d}=10\log_{10}\left ( \frac{I}{I_{0}} \right )$

$\therefore I=I_{0}\times10^{\frac{\beta _{d}}{10}}=10^{-12}\times 10^{60/10}=10^{-6}\ W/m^{2}=1\ \mu W/m^{2}$

Therefore, the total noise power emitted by the baby bat will be

$P=A I=\left (4 \pi \times 1^{2} \ m^{2}\right )\times 1\ \mu W/m^{2}=12.566\ \mu W$

(c) Intensity is inversely proportional to the distance

$I\propto \frac{1}{r^{2}},\ or\ \frac{I_{1}}{I_{2}}=\left (\frac{r_{2}}{r_{1}} \right )^{2} \\ \\ \therefore \ r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}$

$\therefore \ r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}=r_{1}\sqrt{\frac{I_{0}\times10^{\frac{\beta _{d1}}{10}}}{I_{0}\times10^{\frac{\beta _{d2}}{10}}}}=r_{1}\sqrt{\frac{10^{\frac{\beta _{d1}}{10}}}{10^{\frac{\beta _{d2}}{10}}}}$

$\therefore \ r_{2}=1\ m\times \sqrt{\frac{10^{\frac{60}{10}}}{10^{\frac{30}{10}}}}=31.62\ m$

The baby is at a distance of 31.62 m.