A bat flying at 5.0 m/s emits a chirp at 35 kHz. If this sound pulse is reflected by a wall, what is the frequency of the echo received by the bat?

I keep getting the wrong answer, don't understand.

The answer depends upon whether the bat is flying towards the wall or away from it (or parallel to it). The problem should have stated which it is.

If flying away or towards the wall, there are TWO Doppler shifts involved. The wall "receives" and reflects a Doppler-shifted sound, and when the bat hears it, it is shifted again.

Do you need a numerical answer? If the bat is flying towards the wall,

f' = f*[V+Vs) / V]
f' = unknown, f = 35 kHz, V = 343 m/s (in dry air, 20 deg C), Vs = (5.0*2 = 10 m/s) <--Based on drwls</> suggestion, the source moves with a velocity of 5.0 m/s toward the wall, but then the bat continues to move at 5.0 m/s toward the reflected waves, doubling the speed between bat the source</> and bat the observer to 10 m/s </>.

If the bat is flying away from the wall,
f' = f*[V-Vs) / V]. Use the same values as above.

To find the frequency of the echo received by the bat, you need to consider the Doppler effect. The Doppler effect is the change in frequency of a wave in relation to an observer moving relative to the source of the wave.

In this case, the bat is the source of the sound wave, and it is both emitting the chirp and receiving the echo. The bat is moving towards the wall, so the frequency of the sound wave will appear higher to the wall than it actually is. The wall reflects the wave back to the bat, while the bat is still moving towards the wall.

The formula to calculate the shifted frequency due to the Doppler effect in this case is:

f' = f * (v + v₀) / (v - v₀)

Where:
f' = frequency observed by the wall (received echo frequency)
f = original frequency emitted by the bat (35 kHz = 35,000 Hz)
v = velocity of sound in air (approximately 343 m/s at room temperature)
v₀ = velocity of the bat (5.0 m/s)

Plugging in the values:

f' = (35,000 Hz) * (343 m/s + 5.0 m/s) / (343 m/s - 5.0 m/s)

Simplifying:

f' = (35,000 Hz) * (348 m/s) / (338 m/s)

f' ≈ 36,086 Hz

So, the frequency of the echo received by the bat is approximately 36,086 Hz.

To understand why you are getting the wrong answer, let me explain the concept of the Doppler effect in this scenario.

The Doppler effect describes the change in frequency of a wave (in this case, sound) due to the relative motion between the source of the wave and the observer. When the source of the wave moves towards the observer, the frequency appears higher, and when it moves away, the frequency appears lower.

In this case, the bat is emitting a chirp at 35 kHz while flying towards a wall. The sound wave from the chirp will be compressed (higher frequency) as the bat moves closer to the wall, and then reflected back towards the bat. As the bat moves away from the wall, the sound wave will be stretched (lower frequency). Therefore, the frequency of the echo received by the bat will be different from the original frequency of the chirp.

To calculate the frequency of the echo received by the bat, you need to consider the relative velocity between the bat and the wall. Since the bat is the source of the chirp and it is moving towards the wall at 5.0 m/s, the relative velocity is the sum of their velocities. The relative velocity is positive because the bat is moving towards the wall.

Now, let's calculate the frequency of the echo received by the bat.

1. Calculate the speed of sound in air:
The speed of sound in air is approximately 343 m/s. This value may vary depending on factors like temperature and humidity, but for this calculation, we'll use the common value of 343 m/s.

2. Calculate the observed frequency:
To determine the observed frequency, we can use the formula for the Doppler effect with the following variables:
- v is the velocity of the bat (5.0 m/s).
- Vs is the speed of sound in air (343 m/s).
- f is the original frequency of the chirp (35 kHz).

The formula for the observed frequency is:
f' = f * (Vs + v) / (Vs)

Plugging in the values:
f' = 35,000 Hz * (343 m/s + 5.0 m/s) / (343 m/s)

Calculating:
f' = 35,000 Hz * 348 m/s / 343 m/s

Simplifying:
f' ≈ 35,489 Hz

Therefore, the frequency of the echo received by the bat is approximately 35,489 Hz.

By using the formula and plugging in the correct values for velocity and speed of sound, you should be able to get the correct answer.