Doppler shift for Bats

DOPPLER SHIFT FOR BATS

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 Bat is hunting for food and is emitting ultrasonic waves in a linear path. The sound waves hit the prey and is reflected back which is detected by the Bat. So the prey is stationary but the sound waves reflected from the prey are moving toward the Bat. Consider a Bat and its prey, the bat is moving toward its prey at a speed of 99 mph [158.4 Kmph]. Initially the Bat is the source but later it is the observer since it’ll receive its own reflected sound from the prey which is the source. We’ll determine the apparent frequency as registered by the Bat when it hears its own reflected sound.

ASSUMPTIONS

1. The atmospheric air 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 humidity on sound is negligible

3. The amplitude of sound is unity

4. The air 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 air at standard temperature and pressure conditions (m/s) {V = 343 m/s}

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

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

The Doppler Effect 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 toward the source while the ‘–’ sign in the denominator indicates that the source is moving toward the observer.

The velocity of source [Prey] V_s = 0 m/s {⸪ Source is stationary} (Eq. 3)

The velocity of observer [Bat] V_o = 99 mph

                                                       = 44 m/s (Eq. 4)

Frequency of Bat sound f₀ = 40000 Hz (Eq. 5)

Speed of sound in air V = 343 m/s (Eq. 6)

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

f’ = 40000*{[343 + 44]/[343]}

f’ = 45131.19 Hz

This is the frequency of sound as registered by the observer [Bat] when sound waves reflected from Prey approach the Bat.

Difference in frequency = f’ – f₀

                                       = 45131.19 – 40000

                                       = 5131.19 Hz

CONCLUSION

We thus determined the apparent frequency as registered by the Bat due to Doppler shift. The Bat must consider this effect in order to catch its prey. At first it may not be aware of the Doppler shift hence may end up not accurately locating the prey. But with experience, it will be able to judge the position of prey based on the fact that emitted and received frequency aren’t the same.