What Speed Is the Bat Catching Up to Its Prey?

[jrrodri7]

Homework Statement

Assuming the speed of sound is 338 m/s:

A bat moving at 6.2 m/s is chasing a flying insect. The bat emits a 53 KHz chirp and receives an echo at 53.6 KHz. At what speed is the bat catching up on its prey?

Homework Equations

$$f^{'}$$=[ ($$V_{Air}$$ + $$V_{observer}$$) $$/$$ ($$V_{Air}$$ - $$V_{source}$$ ] * f

The Attempt at a Solution

I assumed that since the objects are moving in the same direction I said that the signs are different, but I don't know which one should be positive or negative because normally the signs are the same because Doppler effect usually comes into play when the objects are moving towards or away from each other, but these are moving in the same direction, while the bat is moving quicker and accelerating towards the prey...I just plugged in the numbers, but I don't know which way to work with it...Help?

My answers for Vo (+) and Vs (-) was...Vo = 2.4 m/s
With the Vo (-) and Vs (+) I arrived at...Vo = 10.096

 

[alphysicist]

Hi jrrodri7,

Homework Statement

Assuming the speed of sound is 338 m/s:

A bat moving at 6.2 m/s is chasing a flying insect. The bat emits a 53 KHz chirp and receives an echo at 53.6 KHz. At what speed is the bat catching up on its prey?

Homework Equations

$$f^{'}$$=[ ($$V_{Air}$$ + $$V_{observer}$$) $$/$$ ($$V_{Air}$$ - $$V_{source}$$ ] * f

The Attempt at a Solution

I assumed that since the objects are moving in the same direction I said that the signs are different, but I don't know which one should be positive or negative because normally the signs are the same because Doppler effect usually comes into play when the objects are moving towards or away from each other, but these are moving in the same direction, while the bat is moving quicker and accelerating towards the prey...I just plugged in the numbers, but I don't know which way to work with it...Help?

I don't believe that is correct. If they are both moving towards each other, then the signs in numerator and denominator should be different.

When the observer is moving towards the source, the observed frequency is greater, so in that case the numerator has a positive sign.

When the source is moving towards the observer, the observed frequency is greater, so the denominator has the negative sign.

So in this problem, which would it be?

Also, they are talking about the reflected wave, so you'll need to use the formula twice. What do you get?

My answers for Vo (+) and Vs (-) was...Vo = 2.4 m/s
With the Vo (-) and Vs (+) I arrived at...Vo = 10.096

 

[baba89]

m/s

I would approach this problem by first clarifying the assumptions and given information. The problem states that the bat is moving at a constant speed of 6.2 m/s and emitting a 53 KHz chirp. It also receives an echo at 53.6 KHz. The speed of sound in this scenario is given as 338 m/s.

Next, I would use the Doppler effect equation to calculate the relative velocity between the bat and the prey. However, since the objects are moving in the same direction, we need to consider the signs of the velocities. The bat is moving faster than the prey, so its velocity will be positive. The prey is moving slower, so its velocity will be negative.

Plugging in the given values, we get:

53.6 KHz = [ (338 m/s + Vo) / (338 m/s - (-6.2 m/s)) ] * 53 KHz

Solving for Vo, we get:

Vo = 10 m/s

This means that the bat is catching up on its prey at a speed of 10 m/s.

In conclusion, the bat is catching up on its prey at a speed of 10 m/s, assuming the given values and assumptions are correct. It is important to clarify any uncertainties in the given information and to carefully consider the signs and directions of the velocities in a Doppler effect problem.