A. 90.1 m
The wavelength of a wave is given by:
![\lambda=\frac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bv%7D%7Bf%7D)
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
![\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B1531%20m%2Fs%7D%7B17.0%20Hz%7D%3D90.1%20m)
B. 102 kHz
We can re-arrange the same equation used previously to solve for the frequency, f:
![f=\frac{v}{\lambda}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7Bv%7D%7B%5Clambda%7D)
where for the dolphin:
v = 1531 m/s is the wave speed
is the wavelength
Substituting into the equation,
![f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1531%20m%2Fs%7D%7B0.015%20m%7D%3D1.02%20%5Ccdot%2010%5E5%20Hz%3D102%20kHz)
C. 13.6 m
Again, the wavelength is given by:
![\lambda=\frac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bv%7D%7Bf%7D)
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,
![\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B340%20m%2Fs%7D%7B25.0%20Hz%7D%3D13.6%20m)
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):
![\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B340%20m%2Fs%7D%7B39.0%20Hz%7D%3D8.7%20m)
- Wavelength corresponding to the maximum frequency (f=78.0 Hz):
![\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B340%20m%2Fs%7D%7B78.0%20Hz%7D%3D4.4%20m)
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
![\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B1%7D%7B4%7D%281.00%20mm%29%3D0.25%20mm%3D2.5%5Ccdot%2010%5E%7B-4%7D%20m)
And since the speed of the sound wave is
v = 1550 m/s
The frequency will be
![f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7Bv%7D%7B%5Clambda%7D%3D%5Cfrac%7B1550%20m%2Fs%7D%7B2.5%5Ccdot%2010%5E%7B-4%7D%20m%7D%3D6.2%5Ccdot%2010%5E6%20Hz%3D6.2%20MHz)