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Helga [31]
3 years ago
9

A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only i

f the sharks are separated by 3.60 meters, one being that much farther away than the other. (a) If the ultrasound has a frequency of 105 kHz, show this ability is not limited by its wavelength m < 3.50 m (b) If this ability is due to the dolphin's ability to detect the arrival times of echoes, what is the minimum time difference (in ms) the dolphin can perceive? ms
Physics
1 answer:
Vlad1618 [11]3 years ago
6 0

Answer:

a) Wavelength of the ultrasound wave = 0.0143 m <<< 3.5m, hence its ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = Period of oscillation = 0.00952 ms

Explanation:

The frequency of the ultrasound wave = 105 KHz = 105000 Hz. The speed of ultrasound waves in water ≈ 1500 m/s. Wavelength = ?

v = fλ

λ = v/f = 1500/105000 = 0.0143 m <<< 3.5m

This value, 0.0143m is way less than the 3.5m presented in the question, hence, this ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = The period of oscillation = 1/f = 1/105000 = 0.00000952s = 0.00952 ms

Hope this helps!

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b) y(x,t)=2sin(2\pi( x-1.25 t)  

c) y(3,10)=1.73 m    

Explanation:

a) The speed of a wave is given by the following equation:

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λ is the wavelength

f is the frequency

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b) The harmonic wave has the following equation:

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A is the amplitude (2 m)

k is the wavenumber (2π/λ)  

ω is the angular frequency (2πf)

y(x,t)=2sin(2\pi x-2\pi*1.25 t)  

y(x,t)=2sin(2\pi( x-1.25 t)  

c) Here we need to find the heigth at x=3 m and t =10 s, so we need to find y(3,10).

y(3,10)=2sin(2\pi(3-1.25*10)

y(3,10)=2sin(2\pi(3-1.25*10)              

y(3,10)=1.73 m              

I hope it helps you!

 

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