Doppler effect and concept of Beat Problem

In summary, when a cyclist with a bell ringing at a frequency of 658.8 Hz drives towards a wall with a speed of 3.18 ms-1, the observer hears beats at a frequency of 12.4 Hz due to the bell itself and the reflection of sound from the wall. This is because the wall, as a moving observer, "observes" sound from the cyclist at a higher frequency, which is then reflected back with the modified frequency and heard by the cyclist.
  • #1
MMONISM
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Homework Statement


A cyclist with a bell ringing with a frequency of 658.8 Hz drives towards a wall with a speed of 3.18 ms-1. Just before colliding with the wall the cyclist hears beats, due to the bell itself and the reflection of the sound from the wall. What is the frequency of beats detected by the observer? Assume that the velocity of sound in air is 343 ms-1.
Cyclist_bell_wall.jpg


Homework Equations


f'=f(c+v0)/(c-vs)
fbeat = f1 - f2


The Attempt at a Solution


f'=f(c+v0)/(c-vs)=f(c)/(c-3.18 )=665 Hz
and I am stuck here, I used fbeat = 665 - 658.8 = 6.2 Hz which is incorrect. Could someone help me please?
 
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  • #2
MMONISM said:

Homework Statement


A cyclist with a bell ringing with a frequency of 658.8 Hz drives towards a wall with a speed of 3.18 ms-1. Just before colliding with the wall the cyclist hears beats, due to the bell itself and the reflection of the sound from the wall. What is the frequency of beats detected by the observer? Assume that the velocity of sound in air is 343 ms-1.View attachment 84371

Homework Equations


f'=f(c+v0)/(c-vs)
fbeat = f1 - f2


The Attempt at a Solution


f'=f(c+v0)/(c-vs)=f(c)/(c-3.18 )=665 Hz
and I am stuck here, I used fbeat = 665 - 658.8 = 6.2 Hz which is incorrect. Could someone help me please?
The wall "observes" sound from the cyclist, a moving source, which is at frequency higher than the original. It reflects sound with the modified frequency, and it is heard by the cyclist, as moving observer.
 
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  • #3
ehild said:
The wall "observes" sound from the cyclist, a moving source, which is at frequency higher than the original. It reflects sound with the modified frequency, and it is heard by the cyclist, as moving observer.
Ok, so f'=f(c+v0)/(c-vs)=f(c)/(c-3.18 )=665 Hz
f''=f'(c+3.18)/(c)=671.2Hz
fbeat = 671.2 - 658.8 = 12.4 Hz
Is this correct now?
 
  • #4
I think it is right.
 
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Related to Doppler effect and concept of Beat Problem

1. What is the Doppler effect?

The Doppler effect is a phenomenon that occurs when there is relative motion between a source of waves and an observer. This results in a change in the frequency of the waves perceived by the observer.

2. How does the Doppler effect work?

The Doppler effect works by changing the wavelength of the waves perceived by the observer. If the source is moving towards the observer, the waves will be compressed, resulting in a higher frequency. If the source is moving away from the observer, the waves will be stretched, resulting in a lower frequency.

3. What is the concept of the beat problem?

The beat problem occurs when two waves of slightly different frequencies are combined. This results in a periodic variation in the amplitude of the resulting wave, known as beats.

4. How is the beat problem related to the Doppler effect?

The beat problem is related to the Doppler effect because the change in frequency of the waves perceived by the observer can result in beats when combined with other waves. This can happen when there are two sources of waves moving at different speeds towards or away from the observer.

5. What are some real-life applications of the Doppler effect and beat problem?

The Doppler effect and beat problem have various practical applications such as in weather radar, medical ultrasound, and police radar guns. They are also used in astronomy to measure the speed and distance of stars and galaxies.

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