Question

Part A - Whale communication.

Answer 1

A) 89.5 m

The wavelength of a wave is given by:

$\lambda=\frac{v}{f}$

where v is the wave speed and f the frequency. For the sound emitted by the whales,

$f=17.0 Hz$

$v=1521 m/s$

Therefore, the wavelength is

$\lambda=\frac{1521 m/s}{17.0 Hz}=89.5 m$

B) 101.4 kHz

The speed of the sound emitted by the dolphins in the water is still

v = 1521 m/s

While the wavelength is

$\lambda=1.50 cm=0.015 m$

So, re-arranging the previous equation we find the frequency:

$f=\frac{v}{\lambda}=\frac{1521 m/s}{0.015 m}=101,400 Hz=101.4 kHz$

C) 1.31 cm

The frequency of the dog whistle is

f = 26.0 kHz = 26,000 Hz

While the speed of sound in air is

v = 340 m/s

Therefore, the wavelength is

$\lambda=\frac{v}{f}=\frac{340 m/s}{26,000 Hz}=0.0131 m=1.31 cm$

D) 4.4 - 8.7 mm

The speed of the sound waves emitted by the bats in air is

v = 340 m/s

The minimum frequency is

f = 39.0 kHz = 39,000 Hz

So the corresponding wavelength is

$\lambda=\frac{v}{f}=\frac{340 m/s}{39,000 Hz}=0.0087 m=8.7 mm$

The maximum frequency is

f = 78.0 kHz = 78,000 Hz

So the corresponding wavelength is

$\lambda=\frac{v}{f}=\frac{340 m/s}{78,000 Hz}=0.0044 m=4.4 mm$

E) 6.2 MHz

The wavelength of the sound must be 1/4 the size of the tumor, so

$\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4}m$

while the speed of sound across the tissue is

v = 1550 m/s

So the frequency must be

$f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz$