Doppler shift underwater Pt7

DOPPLER SHIFT UNDERWATER PT7

INTRODUCTION

We know that Doppler Effect or Doppler shift occurs between a source and observer when they are in relative motion with respect to each other. In this case we’ll determine the Doppler shift that occurs when one dolphin is chasing the other in a linear path. Consider two dolphins, dolphin1 [source] moving at 12 mph chasing dolphin2 [observer] which is moving at 8.3 mph. We’ll determine the apparent frequency as registered by dolphin2 when dolphin1 emits sound waves.

ASSUMPTIONS

1. Water has standard temperature and pressure conditions

·         Temperature T = 298 K or 25°C or 77°F

·         Pressure = 1 bar = 10⁵ N/m²

2. The effect of pressure is negligible

3. The amplitude of sound is unity

4. The water molecules do not move with respect to source and observer

CALCULATION

The equation for Doppler shift is given by,

f’ = f₀*{[V ± V_o]/[V ± V_s]} (Eq. 1) 

f₀ – Original frequency (Hz)

f’ – Apparent or observed frequency (Hz)

V – Velocity of Sound in water at standard temperature and pressure conditions {V = 1531 m/s}

V_o – Velocity of dolphin2 [observer] (m/s)

V_s – Velocity of dolphin1 [source] (m/s)

The Doppler shift equation in this case is,

f’ = f₀*{[V – V_o]/[V – V_s]} (Eq. 2)

The ‘–’ sign in the numerator of equation (2) indicates that the observer is moving away from the source while the ‘–’ sign in the denominator indicates that the source is moving toward the observer.

  

The velocity of dolphin2 V_o = 8.3 mph

                                              = 3.68 m/s (Eq. 3)

The velocity of dolphin1 V_s = 12 mph

                                              = 5.33 m/s (Eq. 4)

Frequency of dolphin1 sound f₀ = 10000 Hz (Eq. 5)

Speed of sound in air V = 1531 m/s (Eq. 6)

Substitute equations (3), (4), (5) and (6) in equation (2),

f’ = 10000*{[1531 – 3.68]/[1531 – 5.33]}

f’ = 10010.81 Hz

This is the frequency of sound as registered by the observer [dolphin2] under water when source [dolphin1] moves toward the observer.

Difference in frequency = f’ – f₀

                                       = 10010.81 – 10000

                                       = 10.81 Hz

We notice that the difference is really small compared to the same scenario in air medium. It is because the source/observer velocity can be considered negligible with respect to speed of sound in water. However in air, the source/observer velocity is considerable with the speed of sound in air. Hence the difference in air would be 48.86 Hz

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift.