DOPPLER SHIFT UNDERWATER PT2

DOPPLER SHIFT UNDERWATER PT2

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 a dolphin is moving away from a stationary submarine underwater. Consider a dolphin emitting ultrasound moving at a speed of 8.3 mph [13.28 kmph], receding away from an observer who is inside a stationary submarine. We’ll determine the apparent frequency of dolphin as registered by the observer in the submarine.

ASSUMPTIONS

Water has standard temperature and pressure conditions

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

·         Pressure = 1 bar = 10⁵ N/m²

The effect of pressure is negligible

The amplitude of sound is unity

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 observer (m/s)

V_s – Velocity of Source [Dolphin] (m/s)

Since the observer is stationary,

V_o = 0 (Eq. 2)

Substitute equation (2) in equation (1),

f’ = f₀*{[V]/[V + V_s]}            (Eq. 3)

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

The velocity of Dolphin V_o = 8.3 mph

                                            = 3.68 m/s (Eq. 4)

Frequency of Dolphin’s sound f₀ = 10000 Hz (Eq. 5)

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

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

f’ = 10000*{1531/[1531 + 3.68]}

f’ = 9976.02 Hz

This is the frequency of sound as registered by the stationary observer in the submarine when a dolphin emitting ultrasound moves away from him.

Difference in frequency = f’ – f₀

                                       = 9976.02 – 10000

                                       = – 23.98 Hz

The negative sign indicates that apparent frequency is less than original but the magnitude is always positive.

CONCLUSION

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