Question

Part A - Whale communication.

Blue whales apparently communicate with each other using sound of frequency 17.0Hz , which can be heard nearly 1000km away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is 1521m/s ?
?1 =
m

Part B - Dolphin clicks.
One type of sound that dolphins emit is a sharp click of wavelength 1.50cm in the ocean. What is the frequency of such clicks?
f =
kHz

Part C - Dog whistles.
One brand of dog whistles claims a frequency of 26.0kHz for its product. What is the wavelength of this sound?
? =
cm

Part D - Bats.
While bats emit a wide variety of sounds, one type emits pulses of sound having a frequency between 39.0kHz and 78.0kHz . What is the range of wavelengths of this sound?
?min,?max =
mm

Part E - Sonograms.
Ultrasound is used to view the interior of the body, much as x rays are utilized. For sharp imagery, the wavelength of the sound should be around one-fourth (or less) the size of the objects to be viewed. Approximately what frequency of sound is needed to produce a clear image of a tumor that is 1.00mm across if the speed of sound in the tissue is 1550m/s ?
f =
MHz

Answer 1

A) 89.5 m

The wavelength of a wave is given by:

`\lambda=(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=(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=(v)/(\lambda)=(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=(v)/(f)=(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=(v)/(f)=(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=(v)/(f)=(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=(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=(v)/(\lambda)=(1550 m/s)/(2.5\cdot 10^(-4) m)=6.2\cdot 10^6 Hz=6.2 MHz`