Reflection of a wave by a rigid boundary

In summary, the inversion of the reflected pulse in a mechanical wave can be explained by Newton's third law of action-reaction. When a crest reaches the end of medium A, the last particle of medium A pulls upwards on the first particle of medium B, and the first particle of medium B pulls downwards on the last particle of medium A. This creates an equal and opposite reaction, resulting in the inversion of the reflected wave. The principle of superposition can also be used to determine the phase of the reflected wave, which should be 180 degrees greater than the incident phase. However, this principle is typically used for two different waves, not the same wave being reflected.
  • #1
Kaushik
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I found this on the internet.

The inversion of the reflected pulse can be explained by returning to our conceptions of the nature of a mechanical wave. When a crest reaches the end of a medium ("medium A"), the last particle of the medium A receives an upward displacement. This particle is attached to the first particle of the other medium ("medium B") on the other side of the boundary. As the last particle of medium A pulls upwards on the first particle of medium B, the first particle of medium B pulls downwards on the last particle of medium A. This is merely Newton's third law of action-reaction. For every action, there is an equal and opposite reaction.
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How does the crest reach the end of the medium? As the other end is fixed there is no way the crest can reach the interface. Isn't it?

My book gave an alternative explanation. It stated that as there is no net displacement at the interface, we can use the principle of superposition to find the phase of the reflected wave (The reflected phase should be 180 more than the incident phase). But my question isn't both the same wave? We use the principle of superposition when there are two different waves. Here, the same wave is just being reflected.

Please correct me
 
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Related to Reflection of a wave by a rigid boundary

1. What is reflection of a wave by a rigid boundary?

Reflection of a wave by a rigid boundary is the phenomenon in which a wave traveling through a medium encounters a boundary that does not allow it to pass through, causing the wave to bounce back in the opposite direction.

2. How does a rigid boundary affect the reflection of a wave?

A rigid boundary does not allow the wave to pass through, causing it to reflect back in the opposite direction with the same speed and frequency as the incident wave. The amplitude of the reflected wave may be different depending on the angle of incidence and the properties of the medium.

3. What is the law of reflection for a wave?

The law of reflection states that the angle of incidence is equal to the angle of reflection, with both angles measured from the normal line (a line perpendicular to the boundary). This law applies to all types of waves, including light, sound, and water waves.

4. How does the angle of incidence affect the reflection of a wave?

The angle of incidence determines the direction in which the reflected wave will travel. If the angle of incidence is perpendicular to the boundary, the wave will reflect back in the opposite direction. As the angle of incidence increases, the angle of reflection also increases, causing the reflected wave to travel in a different direction.

5. What are some real-life examples of reflection of waves by rigid boundaries?

Some common examples of reflection of waves by rigid boundaries include echoes in a canyon, sound reflecting off of walls in a room, and light reflecting off of a mirror. Other examples include seismic waves reflecting off of the Earth's crust and ocean waves reflecting off of a sea wall.

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