What is the correct way to calculate the doppler effect in this scenario?

In summary, the problem involves a bat using ultrasound at 40kHz to locate a stationary insect. The bat is flying directly towards the insect at a speed of 4m/s. The frequency at which the bat receives the ultrasound reflections from the insect is calculated using the formula f = f₀((c+v)/(c-v)), where f₀ is the original frequency, c is the speed of sound, and v is the velocity of the wave. However, the correct answer of 41kHz is obtained by taking into account the doppler effect from both the moving source and the moving observer. The first effect involves a moving source and a stationary observer, while the second effect involves a moving observer and a stationary source. By using the appropriate formulas
  • #1
slaw155
41
0
A bat uses ultrasound at 40kHz to locate a stationary insect. If bat is flying directly towards the insect at speed of 4m/s at what frequency does the bat receive the ultrasound reflections from the insect? Speed of sound = 340m/s.

I went frequency = frequency of source x (velocity of wave+velocity of bat)/(velocity of wave)
So 40 = f x (340+4)/(340).
This however gives me the wrong answer from the textbook answer of frequency = 41kHz.
What have I done wrong?
 
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  • #2
The problem is this: the insect receives the sound from the bat and there is a doppler effect there. Then the insect becomes the source, and the bat receives the sound, and once again we have the doppler effect.
For the first effect, we have a moving source (and stationary observer). The bat, which is the source, moves towards the insect. So, you need to use a different formula to compute the effect; see here- http://hyperphysics.phy-astr.gsu.edu/hbase/sound/dopp.html
After that, you have the observer approaching the stationary source. The insect, which is now the source, is being approached by the bat, now the observer. Here you can use the formula you named above.
Hopefully you can do the algebra stuff, the main problem you had was conceptual I think. It gave me approx 40.95 kHz.
 
Last edited:

What is the Doppler effect of wave?

The Doppler effect of wave is a phenomenon that describes the change in frequency of a wave when the source of the wave is moving relative to the observer. This effect is commonly experienced with sound waves, such as when an ambulance siren changes pitch as it passes by.

What causes the Doppler effect of wave?

The Doppler effect of wave is caused by the relative motion between the source of the wave and the observer. When the source is moving towards the observer, the frequency of the wave is perceived to be higher, and when the source is moving away, the frequency is perceived to be lower.

How does the Doppler effect of wave affect different types of waves?

The Doppler effect of wave affects all types of waves, including sound waves, light waves, and even water waves. However, the effect is most noticeable with sound waves due to their slower speed compared to light and water waves.

How is the Doppler effect of wave used in real life?

The Doppler effect of wave has many practical applications, such as in radar technology, where it is used to determine the speed and direction of moving objects. It is also used in medical ultrasound to measure blood flow and in astronomy to study the motion of celestial objects.

Is the Doppler effect of wave always noticeable?

No, the Doppler effect of wave is not always noticeable. It depends on the speed of the source and the observer, as well as the speed of the wave itself. For example, the Doppler effect is more noticeable with fast-moving objects and low-frequency waves, while it is less noticeable with slow-moving objects and high-frequency waves.

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