Doppler shift underwater Pt6

DOPPLER SHIFT UNDERWATER PT6

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 two submarines are moving away from each other in a linear path. Consider two submarines, sub1 [source] and sub2 [observer] moving at a speed of 40 mph [64 kmph] away from each other. We’ll determine the apparent frequency as registered by the observer in sub2 when sub1 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 sub2 [observer] (m/s)

V_s – Velocity of sub1 [source] (m/s)

The observer and source are moving at the same velocity hence,

V_o = V_s (Eq. 2)

Substitute equation (2) in equation (1),

f’ = f₀*{[V – V_o]/[V + V_o]} (Eq. 3)

The ‘–’ sign in the numerator of equation (3) indicates that the observer is moving away from the source while the ‘+’ sign in the denominator indicates that the source is moving away from the observer.

The velocity of sub1 and sub2 V_o = 40 mph

                                                      = 17.77 m/s (Eq. 4)

Frequency of submarine signal f₀ = 400 Hz (Eq. 5)

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

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

f’ = 400*{[1531 – 17.77]/[1531 + 17.77]}

f’ = 390.82 Hz

This is the frequency of sound as registered by the observer [sub2] under water when source [sub1] is receding away from him.

Difference in frequency = f’ – f₀

                                       = 390.82 – 400

                                       = – 9.18 Hz

The negative sign indicates that apparent frequency is less than original but magnitude is always positive. We also 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 – 39.40 Hz

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift. In fact if both submarines emit sound waves then they both will experience Doppler shift since they act as source and observer at the same time for each other.