A. 90.1 m
The wavelength of a wave is given by:
[tex]\lambda=\frac{v}{f}[/tex]
where
v is the speed of the wave
f is its frequency
For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is
[tex]\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m[/tex]
B. 102 kHz
We can re-arrange the same equation used previously to solve for the frequency, f:
[tex]f=\frac{v}{\lambda}[/tex]
where for the dolphin:
v = 1531 m/s is the wave speed
[tex]\lambda=1.50 cm=0.015 m[/tex] is the wavelength
Substituting into the equation,
[tex]f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz[/tex]
C. 13.6 m
Again, the wavelength is given by:
[tex]\lambda=\frac{v}{f}[/tex]
where
v = 340 m/s is the speed of sound in air
f = 25.0 Hz is the frequency of the whistle
Substituting into the equation,
[tex]\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m[/tex]
D. 4.4-8.7 m
Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:
- Wavelength corresponding to the minimum frequency (f=39.0 Hz):
[tex]\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m[/tex]
- Wavelength corresponding to the maximum frequency (f=78.0 Hz):
[tex]\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m[/tex]
So the range of wavelength is 4.4-8.7 m.
E. 6.2 MHz
In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so
[tex]\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m[/tex]
And since the speed of the sound wave is
v = 1550 m/s
The frequency will be
[tex]f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz[/tex]
