Waves and Sounds - speed of a bat

[FrenchAtticus]

Homework Statement

A bat moving at 4.5 m/s is chasing a flying insect insect. The bat emits a 52 kHz chirp and receives back an echo at 52.75 kHz. At what speed is the bat gaining on its prey? Take the speed of sound in air to be 342 m/s.

Homework Equations

f1 = f ((velocity +/- velocity observer)/(velocity +/- velocity source))

The Attempt at a Solution

To start off, I converted 52 kHz to 52,000 Hz and 52.75 kHz

Then my equation looked like:

52750 = 52000 ((342 - vo)/(342-4.5)

I found vo to be .367788 and subtracted that from 4.5 to find the speed gained, but that doesn't seem to work. Can anyone help?

 

[FrenchAtticus]

Can anybody help me out?

 

[thomate1]

Based on the given information, it seems that the bat is gaining on its prey at a speed of 4.5 m/s. This can be calculated by using the Doppler effect equation, which relates the frequency of a wave to the velocity of the source and the observer.

In this case, the source is the bat and the observer is the insect. The frequency of the emitted chirp (f1) is 52,000 Hz and the frequency of the received echo (f2) is 52,750 Hz. The speed of sound in air (v) is 342 m/s.

Using the Doppler effect equation, we can set up the following equation:

f2 = f1 * (v + vb)/(v + vs)

where vb is the velocity of the bat and vs is the velocity of the insect.

Substituting the given values, we get:

52,750 = 52,000 * (342 + vb)/(342 + vs)

Solving for vb, we get:

vb = (52,750 - 52,000 * vs)/vs

Since the bat is moving at a constant speed of 4.5 m/s, we can set vs = 0 and solve for vb:

vb = (52,750 - 52,000 * 0)/0 = 4.5 m/s

Therefore, the bat is gaining on its prey at a speed of 4.5 m/s.